How can I set all HTML code display except image HTML code
Question
Please see the picture below. I need to set my page like below.
I'm using PHP, to show the HTML code I got success.
<?php
$post = "<p>I like this</p>";
$get_post = htmlspecialchars($post);
?>
and the result will be : <p>I like this</p> <-- All html code display.
But now, if I have an image file,
<img src="logo.png"/>
It won't display the image, but html display.
What I want is, how can I set all HTML code display except image HTML code <img> ?
Please see the image below for example.
Additional Information
1. We only have 1 variable, that's $post
2. Assume $post retrieve from database
Example 1: $post = "<html><p>a</p><img src='logo.png'/></html>";
Example 2: $post = "<img src='logo.png'/>";
Please see the picture above for example.
3 answers
solution1
0 2015-02-18 04:09:23
I think concat will do the thing. Comments are welcomed.
<?php
$post1 = "<p>I like this</p>";
$image= "<img src="logo.png"/>";
$post2= "<p>I like this Too</p>";
$get_post = htmlspecialchars($post1)." ".$image." ".htmlspecialchars($post2);
?>
Update
use htmlspecialchars_decode()
for img src you can visit here or this can solve your question.
solution2
0 2015-02-18 04:20:20
you can check if the string have one img tag, then use if condition to get the output:
$post = "<img src='logo.png'/>";
//$post = '<p>I like this</p>';
if(strpos($post, "<img") != false)
{
echo $post;
}else{
$get_post = htmlspecialchars($post);
echo $get_post;
}
solution3
0 2015-02-18 04:31:56
Still confused of your question for my bad English. Perhaps this helps.
<?php
$post='<img src="image/america.png"/>';
echo $post;
?>
UPDATED :
<?php
$post='<img src="image/america.png"/>';
//$post='<p>Hallo</p>';
$detect=substr($post,0,4);
if ($detect){
echo $post;
}
else{
$get_post = htmlspecialchars($post);
echo $get_post;
}
?>
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