doudongdang4483
2013-07-23 17:09
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如何使用PHP在IF CONDITION中回显图像?

I used the following code to echo image. If slno is equal to 1 the code will echo to output the image. But if slno is greater than 1 (i.e. 2, 3 and so on), the code will just output the $row['slno'] value...but the code does not produce anything. Any help is appreciated.

<?php

$image=$row['picture']; //this is the name of the photo i.e. 1.jpg, 2.jpg and so on.
if ($row['slno']=1) //slno holds numerical data 1, 2, 3, 4,
{
  echo '<img width=50 height=50 src="http://localhost/hl/photo/'.$image.'" />';
}
if($row['slno']>1)
{
  echo $row['slno'];
}
?>
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1条回答 默认 最新

  • dongzongzi0379 2013-07-23 17:10
    已采纳

    You assign the value to your variable. You have to compare them:

    if ($row['slno'] == 1) {
        ...
    } else if($row['slno'] > 1) {
        ...
    }
    
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