Mysqli DELETE QUERY 在 PHP 脚本中不起作用

问题描述

我使用以下代码从表中删除条目我想要做的是检查是否从表中删除了任何值.如果删除了一个值，脚本应该打印成功否则为假.这就是我已经实现了.请帮助

Im using the following code to remove an entry from a table what i want to do is to check if any value was deleted from the table.If one value is deleted,the script should print success else false.This is what i have achieved till now.Please help

<?PHP
    $mysqli = new mysqli("SQLHOST.COM","CLIENT","PASSWORD", "DNAME", 1234);

    if ($mysqli->connect_errno) {
        printf("Connect failed: %s
", $mysqli->connect_error);
        exit();
    }
    else
    {

    printf("cONN Sucees");


    if ($result = $mysqli->query("DELETE FROM ktable WHERE code='value'")) {
        printf("Select returned %d rows.
", $result->num_rows);


     printf($result->num_rows);
        $result->close();
    }

    }
    ?>

推荐答案

你删掉你需要返回的是affected_rows http://www.php.net/manual/en/mysqli.affected-rows.php

what you delete what you need to return is affected_rows http://www.php.net/manual/en/mysqli.affected-rows.php

你需要更换什么

if ($result = $mysqli->query("DELETE FROM ktable WHERE code='value'")) {
    printf("Select returned %d rows.
", $result->num_rows);


    printf($result->num_rows);
    $result->close();
}

工作代码

$value = ""; // Set To any Value
$mysqli = new mysqli ( "SQLHOST.COM", "CLIENT", "PASSWORD", "DNAME", 1234 );
if ($mysqli->connect_errno) {
    printf ( "Connect failed: %s
", $mysqli->connect_error );
    exit ();
} else {
    printf ( "cONN Sucees" );
    if ($mysqli->query (sprintf ( "DELETE FROM ktable WHERE code='%s'", mysqli_real_escape_string ( $mysqli, $value ) ) )) {
        printf ( "Affected Rows  %d rows.
", $mysqli->affected_rows );
    }
}

你应该有一个工作输出

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