Simple PHP strpos function not working, why?

问题

Why isn\'t this standalone code working:

$link = \'https://google.com\';
$unacceptables = array(\'https:\',\'.doc\',\'.pdf\', \'.jpg\', \'.jpeg\', \'.gif\', \'.bmp\', \'.png\');

foreach ($unacceptables as $unacceptable) {
        if (strpos($link, $unacceptable) === true) {
            echo \'Unacceptable Found<br />\';
        } else {
            echo \'Acceptable!<br />\';
        }
}

It\'s printing acceptable every time even though https is contained within the $link variable.

回答1:

When in doubt, read the docs:

[strpos] Returns the numeric position of the first occurrence of needle in the haystack string.

So you want to try something more like:

// ...
if (strpos($link, $unacceptable) !== false) {

Because otherwise strpos is returning a number, and you're looking for a boolean true.

回答2:

strpos() does not return true when it finds a match, it returns the position of the first matching string. Watchout, if the match is a the beginning of the string it will return an index of zero which will compare as equal to false unless you use the === operator.

回答3:

Your failure condition is wrong.

strpos returns false if match is not found, so you need to explicitly check

if (strpos($link, $unacceptable) !== false) {

回答4:

Strpos always return position like you search "httpsL" in your string('https://google.com';) then it return 0th position and PHP evaluate it as false.

please see this link:(Hope its very usefull for you): http://php.net/manual/en/function.strpos.php

回答5:

strpos

function may return Boolean FALSE, but may also return a non-Boolean value which evaluates to FALSE.

So I did like this

if (strpos($link, $unacceptable) !== false) {
    //Code
}

来源：https://stackoverflow.com/questions/4858927/simple-php-strpos-function-not-working-why

标签

php

strpos