Question

我正在尝试更新我需要显示MySql错误字段的代码。该脚本检查MySql，如果结果== 0基于传递3个变量，那么$ outstring将在$ outstring中发送错误。这个$ outstring由jQuery使用data.opp来处理，以确定错误或成功，并在对话框窗口中显示结果。发生的事情是我的所有字段都显示在$ outstring中，我只想显示哪个字段失败。

最好的方法是什么，所以我知道哪个$变量产生0结果并显示MySql的哪个部分失败。我已经发布了处理此输出的相关代码部分。非常感谢。

<?php

$outString = ''; // NEED to do HTML output Formating on SERVER Side
    // Only Return a String to browser NOT an array


    foreach ($array as $box) {
        $box = mysql_real_escape_string($box);
        $sql = "SELECT custref,customer,department,status FROM boxes WHERE custref = '$box' ";
        $sql .= "AND customer = '$company' ";
        $sql .= "AND department = '$dept' ";
        $sql .= "AND status = '1' ";
        $result = mysql_query($sql) or die ('{"opp":"error","box":"' . mysql_error() . '"}');

        // If there are no entries, send output to outstring as error
        if (mysql_num_rows($result) == 0) {

            $outString .= $box . ' ';
            $outString .= $dept . ' ';
            $outString .= '1' . ' ';


    }   // foreach ($array
    }
    if ($outString) { // if there are numbers in this string, then there are error Duplicates
        $error = array('opp' => "error", 'box' => $outString); // box will have the TEXT for the Dialog Text
    // IMPORTANT, have changed JSON output to use an Opperation (opp) to indicate
    // in javascript if there was an error OR not
        $output = json_encode($error);
        echo $output;
        exit();
    }

?>

检查MySql是否显示失败的输入

0 个答案: