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Home/ Questions/Q 6686745
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Editorial Team
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Editorial Team
Asked: May 26, 20262026-05-26T05:12:27+00:00 2026-05-26T05:12:27+00:00

if (phpversion() >= ‘4.3.0’){ $string = mysqli_real_escape_string($string); }else{ $string = mysqli_escape_string($string); } All the

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if (phpversion() >= '4.3.0'){
    $string = mysqli_real_escape_string($string);
}else{
    $string = mysqli_escape_string($string);
}

All the documentation for mysqli_real_escape_string seems to indicate this is a valid bit of code, but I don’t understand why I get this error:

mysqli_real_escape_string() expects exactly 2 parameters, 1 given

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  1. Editorial Team
    Editorial Team
    2026-05-26T05:12:28+00:00Added an answer on May 26, 2026 at 5:12 am

    Documentation says it needs two parameters:

    string mysqli_real_escape_string ( mysqli $link , string $escapestr )
    

    The first one is a link for a mysqli instance, the second one is the string to escape.

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