CODEWITHSUNDEEP
phpif-statementmysqli
Alan001
Alan001

Reputation: 13

PHP: Why does else-block not work?

I have the following code:

<form action="" method="POST">
    <input id="delete" name="delete" placeholder="Name" type="text"></b>
    <input type="submit" name="deleteButton" value="Go!">
</form>

<?php
if (isset($_POST['deleteButton'])) {
    $temp = $_POST['delete'];
    $deleteSQL = "DELETE FROM `berater` WHERE `Nachname` LIKE '$temp' OR `Vorname` LIKE '$temp' OR CONCAT(Vorname,' ', Nachname) LIKE '%$temp%'";
    $searchSQL = "SELECT FROM `berater` WHERE `Nachname` LIKE '$temp' OR `Vorname` LIKE '$temp' OR CONCAT(Vorname,' ', Nachname) LIKE '%$temp%'";
    $search = mysqli_query($db_link, $searchSQL);
    $loeschen = mysqli_query($db_link, $deleteSQL);

    if (!$search) {
        echo "Kein Datensatz gefunden!";
    } else {
        echo "$temp erfolgreich gelöscht.";
    }
}

I am new to PHP and thought the mysqli_query returns false if the name is not found in the database. What am I doing wrong here?

Upvotes: 0

Views: 63

Answers (2)

leandronn
leandronn

Reputation: 171

Try using mysqli_num_rows:

if (mysqli_num_rows($search)==0) {
    echo "Kein Datensatz gefunden!";
} else {
    echo "$temp erfolgreich gelöscht.";
}

Upvotes: 0

kopaty4
kopaty4

Reputation: 2296

According to mysqli_query() documentation:

Returns FALSE on failure. For successful SELECT, SHOW, DESCRIBE or EXPLAIN queries mysqli_query() will return a mysqli_result object. For other successful queries mysqli_query() will return TRUE.

Replace if (!$search) with if (mysqli_num_rows($search) > 0) to check if such row existed.

By the way, this code can not be used in production: it is vulnerable to SQL injection attacks. I advise you to read this article before working with the database: http://php.net/manual/en/security.database.sql-injection.php

Upvotes: 1

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