PHP - strstr() returns false but the $haystack contains the $needle

Question

This is a very strange error, i am trying to fix it without success. I am trying to check if a link contains a string:

$actual_link = "http://$_SERVER[HTTP_HOST]$_SERVER[REQUEST_URI]";
echo $actual_link;

The output is:

http://xxx.xxx.xxx.xxx/plesk-site-preview/***********.com/xxx.xxx.xxx.xxx/

Then:

if(strstr($actual_link,"plesk-site-preview")  ){
echo"<meta name='robots' content='noindex'>";
}

The problem is that strstr return false despite the substring plesk-site-preview is contained in http://xxx.xxx.xxx.xxx/plesk-site-preview/***********.com/xxx.xxx.xxx.xxx/.

How can i fix this error?

EDIT:

I have inserted before if(strstr($actual_link,"plesk-site-preview") ){ the following line for testing purpose:

$actual_link='http://xxx.xxx.xxx.xxx/plesk-site-preview/***********.com/xxx.xxx.xxx.xxx/';

Now the code is working! It seems that the string assigned at the variable $actual_link is lost before the IF statement.

Try `if(strstr($actual_link,"plesk-site-preview") !== FALSE ){` — NineBerry, Sep 15 '18 at 21:41
@NineBerry i have inserted your code but the error remains! Please note that my code is exact and there are not further lines of code between `echo $actual_link;` and the `IF` statement. — Daniele, Sep 15 '18 at 21:53
Did you try checking the result with `!== False` ? That should work — Mawg says reinstate Monica, Nov 12 '18 at 19:09

score 1 · Answer 1 · edited Jun 20 '20 at 09:12

1

The documentaion says

string strstr ( string $haystack , mixed $needle [, bool $before_needle = FALSE ] )

Returns part of haystack string starting from and including the first occurrence of needle to the end of haystack.

Returns the portion of string, or FALSE if needle is not found.

And you code

if(strstr($actual_link,"plesk-site-preview"))

Perhaps it ought to be

if(strstr($actual_link,"plesk-site-preview") != "")

as it returns a string, not a boolean if successful.

Hmm, actually it would be better to

 if(strstr($actual_link,"plesk-site-preview") !== FALSE)

edited Jun 20 '20 at 09:12

Community

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answered Sep 15 '18 at 21:40

Mawg says reinstate Monica

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1

To use strict type checking, use `!==` – NineBerry Sep 15 '18 at 21:44
1

@Mawg I have replaced my code with `if(strstr($actual_link,"plesk-site-preview" != FALSE) )` but the error has not been fixed! Please note also that you have forgotten a `)` in your code. – Daniele Sep 15 '18 at 21:45
@NineBerry is correct, `!=` ought to be `!==`. The point here is that many PHP functions return FLASE for failure and the proper way to test is `!== FALSE`; it's a common pattern – Mawg says reinstate Monica Sep 15 '18 at 21:50
Of course, if you used a debugger, you wouldn't have to ask us this :-) – Mawg says reinstate Monica Sep 15 '18 at 21:51

Maksym Fedorov · Answer 2 · 2018-09-15T21:47:23.090

0

If you need check string presence of a substring in a string then you can use strpos, for example:

if(strpos($actual_link, "plesk-site-preview")){
    echo"<meta name='robots' content='noindex'>";
}

This way is better, because strpos is faster than strstr

edited Sep 15 '18 at 21:47

answered Sep 15 '18 at 21:45

Maksym Fedorov

6,383
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I have tried with your code but the error has not been fixed! – Daniele Sep 15 '18 at 21:47
@Daniele, This condition is correct. You can check it, for example: `if(strpos($actual_link, "plesk-site-preview")){ echo "Substing finded"; }` You should check other code. – Maksym Fedorov Sep 15 '18 at 21:54
1

@Daniele, I was tested it. My code: `$actual_link = "http://84.200.70.56/plesk-site-preview/***********.com/84.200.70.56/"; if(strpos($actual_link,"plesk-site-preview") ){ echo "Substring found"; }` and it works – Maksym Fedorov Sep 15 '18 at 22:04
Try print $actual_link variable before if statement – Maksym Fedorov Sep 15 '18 at 22:17
`strpos($actual_link, "plesk-site-preview") !== FALSE` will solve your problem – Martin Nov 10 '18 at 17:28

PHP - strstr() returns false but the $haystack contains the $needle

2 Answers2