MySQL WHERE query not working

Question

I have problem in this code, I'm taking information from two tables and "where news_id=$dz" is not working. Here is the page which include this code.

Here is my code:

<?php
mysql_query("SET NAMES UTF8");
$result = mysql_query("SELECT dz,title FROM dzeglebi where raioni='ყვარლის მუნიციპალიტეტი' && mxare='kaxeti' ORDER BY title ASC", $db);
$myrow = mysql_fetch_array ($result);

printf(
       "<h2><li>
       <strong><a href='../../dzeglebi.php?id=%s'>%s</a>
       </strong></li></h2>",$myrow["dz"],$myrow["title"]);


function FetchImage($id)
{
    $images=array();
    $x=0;
    $d=mysql_query("select * from `images` where news_id=$dz");
    while($data=mysql_fetch_array($d))
    {
        $images["big"][$x]=$data["image"];
        $images["small"][$x]=$data["small"];
        $x++;
    }   
return $images;
}
function CountImages($id)
{
$d=mysql_query("select * from `images` where news_id=$dz");
return mysql_num_rows($d);  
}
$imgs=FetchImage($id);

 for($i=0;$i<CountImages($id);$i++)
{
echo'

 <img src="../'.$imgs["big"][$i].'" >';
}



?>

You should connect to the $db at first. In functions you cant access $dz. In php inside functions you can access only arguments and local variables. So you should pass $dz az a function argument. (Or access it as a global variable, but passing as an argument is a better way.) Probably there are others issues also... — Lajos Veres
– Lajos Veres, Commented Oct 10, 2013 at 20:05
You have $id parameter in FetchImage($id) and CountImages($id) and using where on where news_id=$dz there should be $id — M Khalid Junaid
– M Khalid Junaid, Commented Oct 10, 2013 at 20:19
I understand you but I want to this that "news_id"=$dz how can I do this? — Sapovnela Gapovninebt
– Sapovnela Gapovninebt, Commented Oct 11, 2013 at 12:33

Iłya Bursov · Accepted Answer · 2013-10-10 20:24:27Z

1

function CountImages($id) { $d=mysql_query("select * from images where news_id=$dz"); - here you should use $id

next problem: - DO NOT USE for($i=0;$i<CountImages($id);$i++) change it to

$len = CountImages($id);
for($i=0;$i<$len;$i++)

because you do not need to perform SQL query for each FOR iteration, or even remove it - as you have $images array, iterate over it

next problem:

$d=mysql_query("select * from `images` where news_id=$id");
return mysql_num_rows($d);

DO NOT USE mysql_num_rows, use SQL count(*) - because it will send only 1 number from mysql to php, instead of whole result set

answered Oct 10, 2013 at 20:24

Iłya Bursov

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Comments

Fabien TheSolution · Accepted Answer · 2013-10-10 20:22:07Z

0

In the 1st query, change && mxare='kaxeti' by AND mxare='kaxeti'. Also, in functions FetchImage and CountImages, change news_id=$dz by news_id=$id

answered Oct 10, 2013 at 20:22

Fabien TheSolution

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