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Hey; I hope I can explain this properly for you to understand.. I have a form with a select option and wanting to have when user selects one of the drop down options it , then submit is clicked that it saves to the selected option... Here is what I think is right ..but could be wrong..if anything would like to know if I am close..???...LOL sample code:

	<html><body><?php	//Select table and insert info to selected table	$select = table("");		if ($select == "xyz")	{		$sql="INSERT INTO xyz ( srnumber, tenant, floor, location, job_description, employee, labour_cost, material_cost, date_twa_sent_to_tenant, date_twa_approved, date_parts_ordered, supplier, date_job_completed, date_billing_given_to_accounting, employee_comments, managers_comments) values ('$_POST[srnumber]','$_POST[tenant]','$_POST[floor]','$_POST[location]','$_POST[job_description]','$_POST[employee]','$_POST[labour_cost]','$_POST[material_cost]','$_POST[date_twa_sent_to_tenant]','$_POST[date_twa_approved]','$_POST[date_parts_ordered]','$_POST[supplier]','$_POST[date_job_completed]','$_POST[date_billing_given_to_accounting]','$_POST[employee_comments]','$_POST[managers_comments]')";	}	elseif ($select == "zyx")	{		$sql="INSERT INTO zyx ( srnumber, tenant, floor, location, job_description, employee, labour_cost, material_cost, date_twa_sent_to_tenant, date_twa_approved, date_parts_ordered, supplier, date_job_completed, date_billing_given_to_accounting, employee_comments, managers_comments) values ('$_POST[srnumber]','$_POST[tenant]','$_POST[floor]','$_POST[location]','$_POST[job_description]','$_POST[employee]','$_POST[labour_cost]','$_POST[material_cost]','$_POST[date_twa_sent_to_tenant]','$_POST[date_twa_approved]','$_POST[date_parts_ordered]','$_POST[supplier]','$_POST[date_job_completed]','$_POST[date_billing_given_to_accounting]','$_POST[employee_comments]','$_POST[managers_comments]')";	}	elseif ($select == "yxz")	{		$sql="INSERT INTO yxz ( srnumber, tenant, floor, location, job_description, employee, labour_cost, material_cost, date_twa_sent_to_tenant, date_twa_approved, date_parts_ordered, supplier, date_job_completed, date_billing_given_to_accounting, employee_comments, managers_comments) values ('$_POST[srnumber]','$_POST[tenant]','$_POST[floor]','$_POST[location]','$_POST[job_description]','$_POST[employee]','$_POST[labour_cost]','$_POST[material_cost]','$_POST[date_twa_sent_to_tenant]','$_POST[date_twa_approved]','$_POST[date_parts_ordered]','$_POST[supplier]','$_POST[date_job_completed]','$_POST[date_billing_given_to_accounting]','$_POST[employee_comments]','$_POST[managers_comments]')";	}	elseif ($select == "yxz1")	{		$sql="INSERT INTO yxz1 ( srnumber, tenant, floor, location, job_description, employee, labour_cost, material_cost, date_twa_sent_to_tenant, date_twa_approved, date_parts_ordered, supplier, date_job_completed, date_billing_given_to_accounting, employee_comments, managers_comments) values ('$_POST[srnumber]','$_POST[tenant]','$_POST[floor]','$_POST[location]','$_POST[job_description]','$_POST[employee]','$_POST[labour_cost]','$_POST[material_cost]','$_POST[date_twa_sent_to_tenant]','$_POST[date_twa_approved]','$_POST[date_parts_ordered]','$_POST[supplier]','$_POST[date_job_completed]','$_POST[date_billing_given_to_accounting]','$_POST[employee_comments]','$_POST[managers_comments]')";	}					endif	{				echo "Sorry no db found";				 }	?></body></html>

ThanksEdited 6:05am ..Just looking script over I think it maybe wrong..the select is called when the location is displayed..so I think it should be like this..not sure though...

//Select table and insert info to selected table	$select from table== field("location");

The selection is in the field " Location" which directs the selected field to be saved to the same table name...Example :If location is " thisdb" then also table name is " thisdb "...so info is sent to that database...

Edited by 2old2learn?
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$_POST[tenant] index of it in associative array should be quoted $_POST['tenant'] else it will take it as CONSTANT first and throw some notice and then php will assume it as a string. but it will already throw a error. same with other associative array.

} endif echo "Sorry no db found"; }
there is proably a bracet miss match and there is no endif there. probably you need 'else' there unless you are not going to add another condition. if there is need to add another condition you need to use iflese thenother than that it is looks fine and will serve the purpose but If table name is too many then rather than writing the query part again and again with if else it will be better to use 'switch case' to check (aka validate) the value for $select then use it in the query in the place of table name. and then executing itrather than that i would go for something like that with less amount of code....assuming $select is an option coming from drop down.in_array() will check that certain value is in an array or not.
$validTable=array('xyz','zyx','xyz1');  // prepare table name which should be considered validif(in_array($select,$vaildTable))  //It will check that $select is a in valid list of validTable array. {		 // table is valid in $select prepare for DB work		 $sql="INSERT INTO $select ( srnumbe.......}else{echo 'No matched table there';}

Edited by birbal
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$_POST[tenant] index of it in associative array should be quoted $_POST['tenant'] else it will take it as CONSTANT first and throw some notice and then php will assume it as a string. but it will already throw a error. same with other associative array.there is proably a bracet miss match and there is no endif there. probably you need 'else' there unless you are not going to add another condition. if there is need to add another condition you need to use iflese thenIf table name is too many then rather than writing the query part again and again with if else it will be better to use 'switch case' to check (aka validate) the value for $select then use it in the query in the place of table name. and then executing itrather than that i would go for like that with less amount of code...something like...assuming $select is an option coming from drop down.in_array() will check that certain value is in an array or not.
$validTable=array('xyz','zyx','xyz1');  // prepare table name which should be considered validif(in_array($select,$vaildTable))  //It will check that $select is a in valid list of validTable array. {		 // table is valid in $select prepare for DB work		 $sql="INSERT INTO $select ( srnumbe.......}else{echo 'No matched table there';}

If you mean $select is in the drop down the answer is no...but xyz,zyx,xyz1 are part of the drop down...that are selected..and once one of those options is select and the rest of the form fields are completed..then user clicks submit..and all data is sent to relevent table...But I like your idea of an array..certainly makes it easier and shorter coding.. Edited by 2old2learn?
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okay here is the code for the select:

<td>Location:</td><td><select name="location" >						<option value="Selection"> Make Selection   </option>						<option value="xyz">  xyz </option>															  <option value="yzx">  yzx </option>						<option value="xzy">  xzy </option>																		  <option value="xyz1">  xyz1 </option></td>

When one of the options is selected and the rest of the form fields is completed then submit is clicked and the data is sent to the option selected table...above is just part of form

Edited by 2old2learn?
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then select data will be in $_POST['location'] or $_GET['location'] depending on your form method post or get relatively . then You should check for $_POST['lcoation'] or $_GET['location'] for valid value and then use it in query if value is valid.

Edited by birbal
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then select data will be in $_POST['location'] or $_GET['location'] depending on your form method post or get relatively . then You should check for $_POST['lcoation'] or $_GET['location'] for valid value and then use it in query if value is valid.
Its $_POST['location'] method..So would it be something like this...??
$Select from table where $_POST['location'];

Edited by 2old2learn?
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Its $_POST['location'] method..So would it be something like this...??
are not you selecting the table name from that drop down?so that it will be like
$select=$_POST['location'];$sql="INSERT INTO $select ( srnumbe.......

here $select will evaluate the table name in the query.and $_POST['location'] is not method it is a super global associative array

Edited by birbal
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are not you selecting the table name from that drop down?so that it will be like
$select=$_POST['location'];$sql="INSERT INTO $select ( srnumbe.......

here $select will evaluate the table name in the query.and $_POST['location'] is not method it is a super global associative array

Yes i am selecting table name...from locationWhat about this??? Is this right then also...
if ($select == "xyz")

or

if ($select =$_POST['location'])

Edited by 2old2learn?
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$select=$_POST['location];if ($select == "xyz")

or

if ($_POST['location'] == "xyz")

will be right. watch for '==' double equal to. there is diffrenec in double equal to and single equal to. double equal to check for condition where as single equal to assighn value.

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$select=$_POST['location];if ($select == "xyz")

or

if ($_POST['location'] == "xyz")

will be right. watch for '==' double equal to. there is diffrenec in double equal to and single equal to. double equal to check for condition where as single equal to assighn value.

okay about double equal signs..so I did sort of manage to get it right..thanks...I went with this...
if($_POST['location'] ="xyz")elseif($_POST['location'] ="yxz")

and so on..until last database is selected...

Edited by 2old2learn?
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like birbal already mentioned, you need to use == when doing comparisons or else after the first if, $_POST['location'] will equal xyz

Edited by thescientist
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like birbal already mentioned, you need to use == when doing comparisons or else after the first if, $_POST['location'] will equal xyz
Oh I miss understood it..so it has to be " == " instead of " = " ...thanks..will make correction..
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yes. one is for assignment, two and three are for comparing.http://www.w3schools.com/php/php_operators.asp
thanks looking at it further it makes alot of sense..the best part is I am starting to understand this a little more and more..at least I think I am...
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  • 2 weeks later...

Sorry even after being given the answers..I didn't do anything testing of it for a long time as a matter fact..was busy with other things...but today started working again on project..and ran into some little errors..

Parse error: syntax error, unexpected '{' in C:\Program Files\xampp\htdocs\tecicc\insert.php on line 49
Below is my if/elseif statements and I get..error message...
   $select=$_POST['location'];      if ( $select= '250y' )   {		$sql="INSERT INTO 250y ( srnumber, tenant, floor, location, job_description, employee, labour_cost, material_cost, date_twa_sent_to_tenant, date_twa_approved, date_parts_ordered, supplier, date_job_completed, date_billing_given_to_accounting, employee_comments, managers_comments) values ('$_POST[srnumber]','$_POST[tenant]','$_POST[floor]','$_POST[location]','$_POST[job_description]','$_POST[employee]','$_POST[labour_cost]','$_POST[material_cost]','$_POST[date_twa_sent_to_tenant]','$_POST[date_twa_approved]','$_POST[date_parts_ordered]','$_POST[supplier]','$_POST[date_job_completed]','$_POST[date_billing_given_to_accounting]','$_POST[employee_comments]','$_POST[managers_comments]')";}   elseif($select == '20q')   {   $sql="INSERT INTO 20q ( srnumber, tenant, floor, location, job_description, employee, labour_cost, material_cost, date_twa_sent_to_tenant, date_twa_approved, date_parts_ordered, supplier, date_job_completed, date_billing_given_to_accounting, employee_comments, managers_comments) values ('$_POST[srnumber]','$_POST[tenant]','$_POST[floor]','$_POST[location]','$_POST[job_description]','$_POST[employee]','$_POST[labour_cost]','$_POST[material_cost]','$_POST[date_twa_sent_to_tenant]','$_POST[date_twa_approved]','$_POST[date_parts_ordered]','$_POST[supplier]','$_POST[date_job_completed]','$_POST[date_billing_given_to_accounting]','$_POST[employee_comments]','$_POST[managers_comments]')";}   elseif($select== '1dundas')   {   $sql="INSERT INTO 1dundas ( srnumber, tenant, floor, location, job_description, employee, labour_cost, material_cost, date_twa_sent_to_tenant, date_twa_approved, date_parts_ordered, supplier, date_job_completed, date_billing_given_to_accounting, employee_comments, managers_comments) values ('$_POST[srnumber]','$_POST[tenant]','$_POST[floor]','$_POST[location]','$_POST[job_description]','$_POST[employee]','$_POST[labour_cost]','$_POST[material_cost]','$_POST[date_twa_sent_to_tenant]','$_POST[date_twa_approved]','$_POST[date_parts_ordered]','$_POST[supplier]','$_POST[date_job_completed]','$_POST[date_billing_given_to_accounting]','$_POST[employee_comments]','$_POST[managers_comments]')";}   else($select=='galleria_mall')   {  <<<<< Line 49 error message..   $sql="INSERT INTO galleria_mall ( srnumber, tenant, floor, location, job_description, employee, labour_cost, material_cost, date_twa_sent_to_tenant, date_twa_approved, date_parts_ordered, supplier, date_job_completed, date_billing_given_to_accounting, employee_comments, managers_comments) values ('$_POST[srnumber]','$_POST[tenant]','$_POST[floor]','$_POST[location]','$_POST[job_description]','$_POST[employee]','$_POST[labour_cost]','$_POST[material_cost]','$_POST[date_twa_sent_to_tenant]','$_POST[date_twa_approved]','$_POST[date_parts_ordered]','$_POST[supplier]','$_POST[date_job_completed]','$_POST[date_billing_given_to_accounting]','$_POST[employee_comments]','$_POST[managers_comments]')";}

I hope I am getting this right...

Edited by 2old2learn?
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do all your curly braces match? the answer should be right around line 49...

Edited by thescientist
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Well, I've looked it over and over I guess I need fresh eyes..to find the "{" error...

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"><html><head><title>insert.php</title><link rel="stylesheet" type="text/css" href="/icc/skins/My_Start_Style.css" /><?phpinclude('config.php');foreach($_POST as $key => $value){  if(empty($_POST[$key])){	$formValid = false;  };};?></head><body><?php$formValid = true;if($formValid){  //$con = mysql_connect("localhost","root",""); // if (!$con){   //die('Could not connect: ' . mysql_error());  // };   //	mysql_select_db("tecicc", $con);   //$select = $_POST['location'];   $validTable=array('250y','20q','1dundas','galleria_mall');      if(in_array($select,$vaildTable))   //if ($_POST['location'] = "20q")   {		$sql="INSERT INTO $select ( srnumber, tenant, floor, location, job_description, employee, labour_cost, material_cost, date_twa_sent_to_tenant, date_twa_approved, date_parts_ordered, supplier, date_job_completed, date_billing_given_to_accounting, employee_comments, managers_comments) values ('$_POST[srnumber]','$_POST[tenant]','$_POST[floor]','$_POST[location]','$_POST[job_description]','$_POST[employee]','$_POST[labour_cost]','$_POST[material_cost]','$_POST[date_twa_sent_to_tenant]','$_POST[date_twa_approved]','$_POST[date_parts_ordered]','$_POST[supplier]','$_POST[date_job_completed]','$_POST[date_billing_given_to_accounting]','$_POST[employee_comments]','$_POST[managers_comments]')";}   if (!mysql_query($sql,$con)){      die('Error: ' . mysql_error());   }  echo "<script type='text/javascript'>alert('1 Record added');</script>";   if (isset($_POST['srnumber'])) {  	include('index_sr.php');  }mysql_close($con);{else}echo "<script type='text/javascript'>alert('Sorry, there was a problem with the form.  Please check all data entries, try again');</script>"; if (isset($_POST['srnumber'])) {  include('index_sr.php'); }}; ?> </body></html>

Edited by 2old2learn?
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this is my selection option set up is this right also..

<td>Location:</td><td><select name="location" >						<option value="Selection"> Make Selection   </option>						<option value="250y">  250Yonge	  </option>						<option value="20q">  20 Queen	  </option>						<option value="1dundas">  1 Dundas	  </option>						<option value="galleria_mall">  Galleria/Mall	  </option></td>

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Well, I've looked it over and over I guess I need fresh eyes..to find the "{" error...
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"><html><head><title>insert.php</title><link rel="stylesheet" type="text/css" href="/icc/skins/My_Start_Style.css" /><?phpinclude('config.php');foreach($_POST as $key => $value){  if(empty($_POST[$key])){	$formValid = false;  };};?></head><body><?php$formValid = true;if($formValid){  //$con = mysql_connect("localhost","root",""); // if (!$con){   //die('Could not connect: ' . mysql_error());  // };										   <<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<< Okay found the " {" error... //mysql_select_db("tecicc", $con);   //$select = $_POST['location'];   $validTable=array('250y','20q','1dundas','galleria_mall');      if(in_array($select,$vaildTable))   //if ($_POST['location'] = "20q")   {		$sql="INSERT INTO $select ( srnumber, tenant, floor, location, job_description, employee, labour_cost, material_cost, date_twa_sent_to_tenant, date_twa_approved, date_parts_ordered, supplier, date_job_completed, date_billing_given_to_accounting, employee_comments, managers_comments) values ('$_POST[srnumber]','$_POST[tenant]','$_POST[floor]','$_POST[location]','$_POST[job_description]','$_POST[employee]','$_POST[labour_cost]','$_POST[material_cost]','$_POST[date_twa_sent_to_tenant]','$_POST[date_twa_approved]','$_POST[date_parts_ordered]','$_POST[supplier]','$_POST[date_job_completed]','$_POST[date_billing_given_to_accounting]','$_POST[employee_comments]','$_POST[managers_comments]')";}   if (!mysql_query($sql,$con)){      die('Error: ' . mysql_error());   }  echo "<script type='text/javascript'>alert('1 Record added');</script>";   if (isset($_POST['srnumber'])) {  	include('index_sr.php');  }mysql_close($con);{else	  <<<<<<< This is where the error is poping up...?????}echo "<script type='text/javascript'>alert('Sorry, there was a problem with the form.  Please check all data entries, try again');</script>"; if (isset($_POST['srnumber'])) {  include('index_sr.php'); }}; ?> </body></html>

Okay found the "{" error..so now I get this error...
Parse error: syntax error, unexpected T_ELSE in C:\Program Files\xampp\htdocs\tecicc\insert.php on line 51
Edited by 2old2learn?
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same deal, make sure all you're if/else statements match. It is often easier to see these kinds of thing when you use consistent indenting, tabbing, spacing, etc.

if($condition == true){  //im gonna do some stuff  //some pretty cool stuff  if($stuffToDoIsCool == true){	//now we're getting somewhere!	//wooo!  }else{	 //i guess I'm not that cool	 //but better than being down there!  };}else{  //it's really lonely down here  //all the cool stuff is happening up there  //:(};

just make sure everything lines up and follow standard conventions when it comes to writing out your code.

Edited by thescientist
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