CODEWITHSUNDEEP
php
user3178022
user3178022

Reputation: 5

Why doesn't "str_replace" work when used inside a function?

I need to skip single quotes in some input data and I wanted to do it within a function (the function should include also other instructions, but I'm not writing them here for the sake of clarity). So I wrote the following function and tested its output inside and outside the function:

function quote_skip($data)
    {
        $data = str_replace("'", "\'", $data);
        echo "Output inside the function quote_skip: ".$data." <br>";
        return $data;
    }
    $test = "l'uomo";
    quote_skip($test);
    echo "Output outside the function quote_skip: ".$test."<br>";

The result is the follwing:

Output inside the function quote_strip: l\'uomo

Output outside the function quote_strip: l'uomo

So what happens is that when I echo the variable outside the function the backslash is not there anymore. Why does this happen? Is there a way to keep the backslash also outside the function?

I only know the basics of php and maybe the answer is very obvious, but I haven't been able to find anything in all the forums I have searched. If anyone has a solution it will be greatly appreciated.

Thank you.

Upvotes: 0

Views: 124

Answers (3)

qwertynl
qwertynl

Reputation: 3933

It will work if you make it so the variable passes by reference:

function quote_skip(&$data) // use the `&`
{
    $data = str_replace("'", "\'", $data);
    echo "Output inside the function quote_skip: ".$data." <br>";
}

Demo: http://phpfiddle.org/lite/code/emr-5ap

Upvotes: 0

Glavić
Glavić

Reputation: 43552

Function is not the problem, your code bellow function is, where you don't echo function output:

$test = "l'uomo";
echo "Output outside the function quote_skip: ".quote_skip($test)."<br>";

Upvotes: 2

phihag
phihag

Reputation: 287825

You ignore the return value of your function in

quote_skip($test);

You want this:

$test = quote_skip($test);

Upvotes: 5

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