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Single Record php

TheCatapult

By TheCatapult
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TheCatapult Member

TheCatapult

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Hello friends, I just want to ask if it possible to have something like this: index.php?id=1 to select the content in ID 1. Here's my code. 

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN"   "http://www.w3.org/TR/html4/loose.dtd"><html xmlns="http://www.w3.org/1999/xhtml" lang="en" xml:lang="en"><head><meta http-equiv="Content-Type" content="text/html; charset=utf-8">	    <title>TCB | Online Streaming | </title>	    <style type="text/css">#div{	 background-color: #669999;					    width: 450px;	    font-family:"Arial", Times, serif;	    text-align:justify;}#news{  width: 450px;  height: 450px;    background-color: #99ffff;  position: absolute; top: 500px; left: 0px;</style></head><body><div id="news"><div><img border="0" src="headline.jpg" alt="Programs" width="451" height="82">	    <?$username="";$password="";$database="";mysql_connect("",$username,$password);@mysql_select_db($database) or die( "Unable to select database");$query="SELECT * FROM news ORDER BY id DESC LIMIT 3";$result=mysql_query($query);$num=mysql_numrows($result);mysql_close();?><?$i=0;while ($i < $num) {$bck = ($i % 2 == 0 ) ? " style=\"background:#A3FFA3\"" : " style=\"background: #B7B7E8;\"";$id=mysql_result($result,$i,"id");$title=mysql_result($result,$i,"title");$content=mysql_result($result,$i,"content");$poster=mysql_result($result,$i,"poster");?><div id="div" <? echo $bck ?>>	   <? echo $id;?><a href="\http://www.mywebsite.com/index.php?id=$id\"><b><? echo $title</a>;?></b></a> <br>$i++;}			    ?></div></div>[/body]

But the link does not work, too.Parse error: syntax error, unexpected '/' in /srv/disk4/741839/www/catapultphpmysql.eu.pn/testwebsite/index.html on line 144 Now pls help me to have ?id=N Thanks!

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birbal Devoted Member

birbal

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in that page you need to catch the $_GET['id'] and have to pass it in WHERE clause with select statement to catch appropiate data from database. after that you can fetch the results.

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TheCatapult Member

TheCatapult

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<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN"   "http://www.w3.org/TR/html4/loose.dtd"><html xmlns="http://www.w3.org/1999/xhtml" lang="en" xml:lang="en"><head><meta http-equiv="Content-Type" content="text/html; charset=utf-8">			<title>TCB | Online Streaming | </title>			<style type="text/css">#div{	background-color: #669999;											width: 450px;			font-family:"Arial", Times, serif;			text-align:justify;}#news{  width: 450px;  height: 450px;	background-color: #99ffff;  position: absolute; top: 500px; left: 0px;</style></head><body><div id="news"><div><img border="0" src="headline.jpg" alt="Programs" width="451" height="82">			<?$username="";$password="";$database="";mysql_connect("",$username,$password);@mysql_select_db($database) or die( "Unable to select database");$query="SELECT * FROM news ORDER BY id DESC LIMIT 3";$result=mysql_query($query);$num=mysql_numrows($result);mysql_close();?><?$username="741839_test";$password="building10";$database="741839_test";$_GET['id']mysql_connect("fdb3.eu.pn",$username,$password);@mysql_select_db($database) or die( "Unable to select database");$query="SELECT * FROM news ORDER BY id DESC LIMIT 3 WHERE id='id'";$result=mysql_query($query);$num=mysql_numrows($result);mysql_close();?><div id="div" <? echo $bck ?>>		   <? echo $id;?><a href="\http://www.mywebsite.com/index.php?id=$id\"><b><? echo $title</a>;?></b></a> <br>$i++;}							?></div></div>[/body]

Not work. Parse error: syntax error, unexpected T_STRING in /srv/disk4/741839/www/catapultphpmysql.eu.pn/testwebsite/index.html on line 131

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TheCatapult Member

TheCatapult

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Sorry for the code. It's not the code, actually. Here's the code. I have change but please help me because it still has an error on it.

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN"   "http://www.w3.org/TR/html4/loose.dtd"><html xmlns="http://www.w3.org/1999/xhtml" lang="en" xml:lang="en"><head><meta http-equiv="Content-Type" content="text/html; charset=utf-8">	    <title>TCB | Online Streaming | </title>	    <style type="text/css">#div{	 background-color: #669999;					    width: 450px;	    font-family:"Arial", Times, serif;	    text-align:justify;	    a:link {text-decoration:none;}    /* unvisited link */a:visited {text-decoration:none;} /* visited link */a:hover {text-decoration:underline;}   /* mouse over link */a:active {text-decoration:underline;}  /* selected link */}#news{a:link {text-decoration:none;}    /* unvisited link */a:visited {text-decoration:none;} /* visited link */a:hover {text-decoration:underline;}   /* mouse over link */a:active {text-decoration:underline;}  /* selected link */  width: 450px;  height: 450px;    background-color: #99ffff;  position: absolute; top: 500px; left: 0px;}</style></head><body><div id="news"><div><img border="0" src="headline.jpg" alt="Programs" width="451" height="82">	    <?$username="";$password="";$database="";$id=mysql_result($result,$i,"id");mysql_connect("",$username,$password);@mysql_select_db($database) or die( "Unable to select database");$query="SELECT * FROM news ORDER BY id DESC LIMIT 3 WHERE id='$id'";$result=mysql_query($query);$num=mysql_numrows($result);mysql_close();?><?$i=0;while ($i < $num) {$bck = ($i % 2 == 0 ) ? " style=\"background:#A3FFA3\"" : " style=\"background: #B7B7E8;\"";$id=mysql_result($result,$i,"id");$title=mysql_result($result,$i,"title");$content=mysql_result($result,$i,"content");$poster=mysql_result($result,$i,"poster");?></body></html>

Thanks!

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justsomeguy w3schools Guru

justsomeguy

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I'm not going to just tell you the answer, but I'll help you figure it out. Look at the error message you posted, it says unexpected end. That means the code ended when the parser was not expecting it to. That means that you are missing a closing bracket in your code. Look at your code carefully and you'll see that you have an opening bracket without a closing one, so you need to add a closing bracket in the place that makes sense for what you're trying to do.

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TheCatapult Member

TheCatapult

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I found it. 

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN"   "http://www.w3.org/TR/html4/loose.dtd"><html xmlns="http://www.w3.org/1999/xhtml" lang="en" xml:lang="en"><head><meta http-equiv="Content-Type" content="text/html; charset=utf-8">	    <title>TCB | Online Streaming | </title>	    <style type="text/css">#div{	 background-color: #669999;					    width: 450px;	    font-family:"Arial", Times, serif;	    text-align:justify;	    a:link {text-decoration:none;}    /* unvisited link */a:visited {text-decoration:none;} /* visited link */a:hover {text-decoration:underline;}   /* mouse over link */a:active {text-decoration:underline;}  /* selected link */}#news{a:link {text-decoration:none;}    /* unvisited link */a:visited {text-decoration:none;} /* visited link */a:hover {text-decoration:underline;}   /* mouse over link */a:active {text-decoration:underline;}  /* selected link */  width: 450px;  height: 450px;    background-color: #99ffff;  position: absolute; top: 500px; left: 0px;}</style></head><body><div id="news"><div><img border="0" src="headline.jpg" alt="Programs" width="451" height="82">	    <?$username="";$password="";$database="";$i=0;while ($i < $num) {}$id=mysql_result($result,$i,"id");mysql_connect("",$username,$password);@mysql_select_db($database) or die( "Unable to select database");$query="SELECT * FROM news ORDER BY id DESC LIMIT 3 WHERE id='$id'";$result=mysql_query($query);$num=mysql_numrows($result);mysql_close();?><?$i=0;while ($i < $num) {}$bck = ($i % 2 == 0 ) ? " style=\"background:#A3FFA3\"" : " style=\"background: #B7B7E8;\"";$id=mysql_result($result,$i,"id");$title=mysql_result($result,$i,"title");$content=mysql_result($result,$i,"content");$poster=mysql_result($result,$i,"poster");?></body></html>

The problems were: Warning: mysql_numrows(): supplied argument is not a valid MySQL result resource in /srv/disk4/741839/www/catapultphpmysql.eu.pn/testwebsite/news.php on line 43Warning: mysql_result(): supplied argument is not a valid MySQL result resource in /srv/disk4/741839/www/catapultphpmysql.eu.pn/testwebsite/news.php on line 53Warning: mysql_result(): supplied argument is not a valid MySQL result resource in /srv/disk4/741839/www/catapultphpmysql.eu.pn/testwebsite/news.php on line 54Warning: mysql_result(): supplied argument is not a valid MySQL result resource in /srv/disk4/741839/www/catapultphpmysql.eu.pn/testwebsite/news.php on line 55Warning: mysql_result(): supplied argument is not a valid MySQL result resource in /srv/disk4/741839/www/catapultphpmysql.eu.pn/testwebsite/news.php on line 56 Thank you very much for help!

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thescientist w3schools Guru

thescientist

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that's odd, because that doesn't look it would work (or make sense) at all. or are you saying that those are errors you are still getting and don't know why? If that is the case (as imagine it must) it would be prudent to go through every line of the PHP code and be able to articulate exactly what the line does, and why it's there, so you can better understand what your code is doing (or isn't doing). Why are you using the first while loop? Do you know how/when to use a while loop?What value do you think you are getting with this line?

$id=mysql_result($result,$i,"id")

Do you know how mysql_query works? (specifically its return value?)Do you know how mysql_result works?

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TheCatapult Member

TheCatapult

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I can understand somewhat but not all. mysql_query is to send a unique query. For example mysql_query('SELECT title FROM news');mysql_result is the one which result what mysql_query has queried. For example:$query=mysql_query('SELECT title from news');mysql_result=('$query'); I'm not sure. All I want is to have a single record. http://www.mywebsite.com/news.php?id=1 Thank you very much.

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Ingolme w3schools Guru

Ingolme

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You must understand all of your code. If you just copy code from somewhere and put it on your page you'll never be able to debug it. Do you understand the purpose of curly braces {} or know what $i is for?

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thescientist w3schools Guru

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the wonder of the internet is that you can look up what you don't know, rather than just guessing and not understanding.

I can understand somewhat but not all. mysql_query is to send a unique query. For example mysql_query('SELECT title FROM news');mysql_result is the one which result what mysql_query has queried. For example:$query=mysql_query('SELECT title from news');mysql_result=('$query'); I'm not sure. All I want is to have a single record. http://www.mywebsite.com/news.php?id=1 Thank you very much.
http://www.w3schools.com/php/php_mysql_select.asphttp://www.w3schools.com/php/func_mysql_query.asphttp://www.w3schools.com/php/func_mysql_result.asp
Hi Inglome, Sorry for my slow understanding. {} I do not know. $i, I do not know. Please help me.
http://www.w3schools.com/php/php_looping.asphttp://www.w3schools.com/php/php_variables.asp you should really take a step back and think exactly of what you are trying to do. blindly throwing code at the situation is not going to help you understand why and how programming works. So, write it out in steps, what you are trying to do. Then take each step and write out its equivalent in code. What you are trying to do is very simple, and all you are doing is complicating it by not trying understand it, thus furthering your own confusion.
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TheCatapult Member

TheCatapult

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in that page you need to catch the $_GET['id'] and have to pass it in WHERE clause with select statement to catch appropiate data from database. after that you can fetch the results.
Thanks thescientist, Inglome, justsomeguy, and birbal for your help. I'm just confused. I think I have done this before but I can't remember the procedure, selecting content by ID. Where will I put the select statement?
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astralaaron Dedicated Member

astralaaron

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I'm just confused. I think I have done this before but I can't remember the procedure, selecting content by ID.Where will I put the select statement?
I am not sure exactly what is confusing you. But the PHP pages read the code from top to bottom, so as long as you don't try to catch the id below the SQL then you are good.. If you do your select statement just like you have but just add in WHERE `id_field_name`=$id_variable it should work out
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TheCatapult Member

TheCatapult

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Hi Inglome, I have searched and have read a few tutorials. I've been reading an article here: http://www.freewebmasterhelp.com/tutorials/phpmysql/6 To quote: Selecting A Single RecordAt the end of the last part of this tutorial, I showed you how to select records from the database based on the contents of partiular fields using: SELECT * FROM contacts WHERE field='value'Now, by using the unique ID field we can select any record from our database using: SELECT * FROM contacts WHERE id='$id'Where $id is a variable holding a number of a record. This may seem to be a little worthless as it is, but you can use this very effectively in a number of different ways. For example, if you wanted to have a dynamically generated site run through a database and a single PHP script, you could write the script to include the database data into the design. Then, using the id field, you could select each individual page and put it into the output. You can even use the page's URL to specify the record you want e.g. http://www.yoursite.com/news/items.php?item=7393And then have the PHP script look up the record with the id corresponding to $item, which in this case would be 7393 I really can't understand this. It has a variable $id but the tutorial does not say what must be on that variable. Or would that variable contain mysql_select ('SELECT * FROM id'). Thanks thanks and many thanks!

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