如何使用PHP在IF CONDITION中回显图像？
问题描述：
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 $row['slno'];
}
?>
答
You assign the value to your variable. You have to compare them:
if ($row['slno'] == 1) {
...
} else if($row['slno'] > 1) {
...
}