Problem with this code is not executed with php

[sonita]

hello Is there something like this in your website.w3schools Fourm -->Server Scripting -->PHP I wish I had something like this on my website. and I have written the code as follows:

• 	<?php	function DisplayPath($cat_id){	mysql_connect('localhost','keyvandb','') or die('connection error');	mysql_select_db('keyvandb') or die('Database error');	mysql_query('SET NAMES\'utf8\'');	mysql_set_charset('utf8');	$cat_id=mysql_real_escape_string($cat_id);	$category=query("SELECT * FROM 'categories' WHERE('id'='{cat_id}') LIMIT 1");	$nav='';	if($category && mysql_num_rows($category)>0){	$category=mysql_fetch_assoc($category);	do {	$nav= '<a herf=" '.ADDR.'/categories/'.$category['id'].'/'.$category['name'].' ">'.$category['name'] .'</a> -> '.$nav;	$parent=mysql_real_escape_string($category['parent']);	$category =query("SELECT * FROM 'categories' WHERE('id'='{$parent}') LIMIT 1");	if($category && mysql_num_rows($category)>0){	$category=mysql_fetch_assoc($category);	}	} while (is_array($category));	//remove the last ->	$nav=mb_substr($nav,0,mb_strlen($nav,'utf-8')-7,'utf-8');	}	echo '<b>دسته بندی:</b>'.$nav.PHP_EOL();	}	?>	

  but don`t exe? why? and i save myinformation in DataBase . please help me?

[justsomeguy]

What happens when you run that? Are there error messages? Possible problems I see are things like not having a space after SET NAMES, calling a function called query when you're using mysql_query in other places, and not checking for errors when running the queries.

[sonita]

nothing error, only my page without this iswhy?how do i? please help me ,very very important.

[justsomeguy]

It might be because of the issues I mentioned above. Add this code to the top of the function to make sure error messages are enabled: ini_set('display_errors', 1);error_reporting(E_ALL);

[sonita]

I add that code(you say) to the top of the function but show nothing error or message .how do i now? :(I put a code sample that I might test runs,please

[justsomeguy]

Going back to what I said in post 2, first put a space in the SET NAMES query. Change your query functions to mysql_query, I doubt you have a function called query defined.

[sonita]

I put a space in the SET NAMES & again run but don`t show.i write this query because :I am iranian & my language is presion ,So it should accept my language & utf8`s language . and show they.but i don`t why change this query? how?What do I have to run this code? What changes can I add to it? Until I see the results on my website.please help me & explain with example for me.

Edited May 7, 2013 by sonita

[birbal]

I suspect you are not calling the function displayPath(). are you calling it somewhere? if it does not get called it wont execute so it wont show any runtime error or do anything.

[justsomeguy]

Look, here you use a function called mysql_query: mysql_query('SET NAMES \'utf8\''); But here, you use a function called query: $category=query("SELECT * FROM 'categories' WHERE('id'='{cat_id}') LIMIT 1"); The mysql_query function is built-in, here it is: http://www.php.net/manual/en/function.mysql-query.php But there's no function called query: http://www.php.net/manual-lookup.php?pattern=query&scope=quickref So, did you create a function called query? If not, then it's going to be an error to try and use a function that doesn't exist. That's what I'm trying to tell you.

[sonita]

helloi change above code as follows:this is code:

<?php require_once 'config.php'; ?><?phpfunction DisplayPath($cat_id) {    error_reporting(E_ALL);    mysql_connect('localhost', 'root', '') or die('Connection error');    mysql_select_db('keivandb') or die('Database error');    mysql_query('SET NAMES\'utf8\'');    mysql_set_charset("utf8");    $cat_id = mysql_real_escape_string($cat_id);    $category = mysql_query("SELECT * FROM categories WHERE id={$cat_id} LIMIT 1");    $nav = '';$name="";$id="";$parent="";while($x = mysql_fetch_array($category))    {   	 $name= $x['name'];	 $id=$x['id'];	 $parent=$x['parent'];     }$cat="categories";    if($category && mysql_num_rows($category) > 0) {	    $category = mysql_fetch_assoc($category);	    do {		    //$nav = '<a href=\" ' . ADDR . '/categories/' . $category['id'] . '/' . $category['name'] . ' \">' . $category['name'] . '</a> -> ' . $nav;     $nav = "<a href=\"http://localhost/test1/{$cat}/{$id}/{$name}\" >{$name}<a>";   echo $nav;  		    $parent = mysql_real_escape_string($category['parent']);		    $category = mysql_query("SELECT * FROM categories WHERE id={$parent} LIMIT 1");		    if($category && mysql_num_rows($category) > 0) {			    $category = mysql_fetch_assoc($category);		    }		    else {			    echo mysql_error();		    }	    } while (is_array($category));	    $nav = mb_substr($nav, 0, mb_strlen($nav, 'utf-8') - 7, 'utf-8');    }    else {	    echo mysql_error();    }  echo '<b>دسته بندی:</b>' . $nav;    //echo '<b>دسته بندی:</b>' . $nav . PHP_EOL();}?>

when exe it show this error (error: you have an error in your SQL syntax; check the manual that corresponds to your MYSQL server version for the right syntax to use near 'LIMIT1' at line 1) what`s meaning error? What can I do to avoid error and run this code?please help me.

[terryds]

I think the {$parent} should be with no curly brackets... Correct me if i'm wrong...

[sonita]

thanks alot ,it exe without error.it`s corectthanks terryds ,

[birbal]

I think the {$parent} should be with no curly brackets... Correct me if i'm wrong...

It should work with curly braces too. it is called interpolation, if a variable enclosed in curly braces inside double quotes it will evaluate its value.

[sonita]

hello,i change above code again , this code exe in myweb page but don`t show The w3schools Fourm -->Server Scripting -->PHP and ArrangedBut as messy as follows (in picture box drawn around it and my language is Persian.)servr scripting->php->w3schools(in persian:طرح آجری-->دسته بندی->طرح) why?this is code :

<?php require_once 'config.php'; ?><?phpfunction DisplayPath($cat_id) {	error_reporting(E_ALL);	mysql_connect('localhost', 'root', '') or die('Connection error');	mysql_select_db('keivandb') or die('Database error');	mysql_query('SET NAMES\'utf8\'');	mysql_set_charset("utf8");	$cat_id = mysql_real_escape_string($cat_id);	$category = mysql_query("SELECT * FROM categories WHERE id={$cat_id} LIMIT 1");	$nav = ''; $name="";$id="";$parent=""; while($x = mysql_fetch_array($category))	{  	 $name= $x['name'];	 $id=$x['id'];	 $parent=$x['parent'];			}$category = mysql_query("SELECT * FROM categories WHERE id={$cat_id} LIMIT 1");while($x = mysql_fetch_array($category))	{  	 $name= $x['name'];  	} $cat="categories";	if($category && mysql_num_rows($category) > 0) {		$category = mysql_fetch_assoc($category);		do {			//$nav = '<a href=\" ' . ADDR . '/categories/' . $category['id'] . '/' . $category['name'] . ' \">' . $category['name'] . '</a> -> ' . $nav;     $nav = "<a href=\"http://localhost/Test2/{$cat}/{$id}/{$name}\" >{$name}<a>";   echo $nav;  			$parent = mysql_real_escape_string($category['parent']);			$category = mysql_query("SELECT * FROM categories WHERE id=parent LIMIT 1");			if($category && mysql_num_rows($category) > 0) {				$category = mysql_fetch_assoc($category);			}			else {				echo mysql_error();			}		} while (is_array($category));		$nav = mb_substr($nav, 0, mb_strlen($nav, 'utf-8') - 7, 'utf-8');	}	else {		echo mysql_error();	}  echo '<b>دسته بندی:</b>' . $nav;	//echo '<b>دسته بندی:</b>' . $nav . PHP_EOL();}?>

how i do Properly and regularly runs? please help me.

[sonita]

please help me ,Why does not anyone answer my question.؟

[thescientist]

whats the problem you are having with it now? Are you getting any errors? Are you debugging to find where you might have logic errors? Its hard to know what to help you with when you don't tell us what's wrong.