Selecting Tables

[deved]

Is it possible to select different tables within a database in MySQL based on the input from a form?If so, could anyone advise on what commands are required to make it work with radio buttons.Any help would be much appreciated.

[aspnetguy]

You mean execute a different query based on what radio button is selected???

$radioresult = $_POST['radiobuttonname'];if($radioresult == "something"){    $query = "sql statement";}else if($radioresult == "something else"){    $query = "a different sql statement";}

[deved]

You mean execute a different query based on what radio button is selected???

$radioresult = $_POST['radiobuttonname'];if($radioresult == "something"){    $query = "sql statement";}else if($radioresult == "something else"){    $query = "a different sql statement";}

<{POST_SNAPBACK}>

Sort of, but instead of using the 2 radio buttons to perform a query within the table... actually using the 2 radio buttons to select one of 2 tables, then executing other queries.

[justsomeguy]

It's the same thing as what aspnetguy posted, just a decision structure based off the radio button.

if($radioresult == "something"){   $table = "table1";}else if($radioresult == "something else"){   $table = "table2";}$sql = "SELECT * FROM {$table} WHERE ... ";

[deved]

It's the same thing as what aspnetguy posted, just a decision structure based off the radio button.

if($radioresult == "something"){   $table = "table1";}else if($radioresult == "something else"){   $table = "table2";}$sql = "SELECT * FROM {$table} WHERE ... ";

<{POST_SNAPBACK}>

Thanks for that. Sorry aspnetguy, still a bit new to PHP and sometimes can't work out the code unless it hits me in the face. Cheers.

[deved]

Can 2 tables be specified from the 1 result such as:

if($radioresult == "something"){   $table = "table1a" && $tbl = "table1b";}else if($radioresult == "something else"){   $table = "table2a" && $tbl = "table2b";}

I'm trying to retrieve data from a row of one table by using data in a row from the second table. The link is a column in both tables that has identical data.

[justsomeguy]

Well, the code you wrote is syntactally correct, but it doesn't do what you think. && is a logical AND operator, it evaluates to a boolean true/false. So this statement:$table = "table1a" && $tbl = "table1b";gets parsed like this:$table = ("table1a" && $tbl = "table1b");$table = ("table1a" && ($tbl = "table1b"));Since $tbl = "table1b" is a logical true statement (assigment statements always evaluate to logical true), it goes to this:$table = ("table1a" && true);The statement "table1a" && true also evaluates to true, since any non-empty value (such as "table1a") is considered true.So, in the end, this statement:$table = "table1a" && $tbl = "table1b";has the effect of setting $tbl equal to "table1b" and $table equal to boolean true. If you want to set two different variables, just use two statements:

if($radioresult == "something"){  $table = "table1a";  $tbl = "table1b";}

[deved]

if($radioresult == "something"){  $table = "table1a";  $tbl = "table1b";}

Can u tell me why this doesn't work (I only get the result from the first query).

<?php$con = mysql_connect("localhost","name","password");if (!$con) {die('Could not connect: ' . mysql_error());}mysql_select_db("dBase", $con);$acc = $_POST['acc'];$radio = $_POST['area'];if ($radio == "input1"){$table = "table1a"; $tbl = "table1b";}else if ($radio == "input2"){$table = "table2a"; $tbl = "table2b";}$result = mysql_query("SELECT * FROM {$table} WHERE account='$acc'");while($row = mysql_fetch_array($result)) { echo "<h3>Job Number</h3>"; echo $row['job']; } $result = mysql_query("SELECT * FROM {$tbl} WHERE Job='job'");while($row = mysql_fetch_array($result)) { echo $row['data']; }?>

[justsomeguy]

It looks fine, I would guess that there is nothing in the table where Job='job'. Open up the database and doublecheck.

[deved]

It looks fine, I would guess that there is nothing in the table where Job='job'.  Open up the database and doublecheck.

<{POST_SNAPBACK}>

I checked and there is data in the field.I also entered a lineecho "DATA";aboveecho $row['data'];which didn't show upIn the 2nd $result line I've tried linking Job='job' to the line echo $row['job']; could this be the cause?Could the result at line echo $row['job']; be turned into a variable, which can then be used in the 2nd query?Have just tested and its definately the link between Job='job' and echo $row['job']; by removing the WHERE function, and all the data including heading DATA showed. Edited June 23, 2006 by deved

[justsomeguy]

I'm not sure I'm following. If you still have problems, post your code and let me know.

[deved]

I'm not sure I'm following.  If you still have problems, post your code and let me know.

<{POST_SNAPBACK}>

Problem sorted. Ended up running a more specific query that matched information from both tables.It would be useful however, if information retrieved and echoed, such as the "echo $row['job']; line, could then be used in another query, as in the line $result = mysql_query("SELECT * FROM {$tbl} WHERE Job='job'");. I'm sure there's a way and with more experience will probably be able to work it out.Thanks for all your help to now tho.

while($row = mysql_fetch_array($result)){echo "<h3>Job Number</h3>";echo $row['job'];}$result = mysql_query("SELECT * FROM {$tbl} WHERE Job='job'");while($row = mysql_fetch_array($result)){echo $row['data'];

[justsomeguy]

You can substitute any string into any other string, it doesn't matter if you're using it for databases or output or whatever, it's just a string.$result = mysql_query("SELECT * FROM {$tbl} WHERE Job='{$row['job']}'");

[deved]

You can substitute any string into any other string, it doesn't matter if you're using it for databases or output or whatever, it's just a string.$result = mysql_query("SELECT * FROM {$tbl} WHERE Job='{$row['job']}'");

<{POST_SNAPBACK}>

Would it then be possible to send the result of echo $row['job'] to another page using $_POST without using fields in a form? If that makes any sense.

[justsomeguy]

If it's going through POST, you will need to submit it in a form, but it can be a hidden value so that it doesn't show up on the page. If you don't want to use a form, you can keep it in the session and have access on the next page.

[deved]

If it's going through POST, you will need to submit it in a form, but it can be a hidden value so that it doesn't show up on the page.  If you don't want to use a form, you can keep it in the session and have access on the next page.

<{POST_SNAPBACK}>

How do you set the value to hidden if value has already been used ie. <input type="text" name="name" value="$var">I'm also having trouble finding info on sessions. Could you tell me how far off the mark this is?Insert the <?php session_start(); ?> before the <html> tag (is this required on each page)?Insert$_SESSION['user'] = $user;$_SESSION['pass'] = $encrypted_pass;on the following page somewhere after the $_POST, $user & $encrypted_passSorry for all the questions, but I've tried a number of tutorials and they're a bit vague in this department.

[justsomeguy]

If you have a value, you can do whatever you want with it. You can display it on the page, or put it in a form variable, or whatever you want. There aren't any constraints on what you can do, all you have is data. <input type="text" name="name" value="<?php echo $var ?>">You're just printing a variable, PHP doesn't care where you print it, and the browser doesn't know any difference.For sessions, that's pretty much it. You can see some examples here:http://www.php.net/manual/en/function.session-start.php