php - 如何为多行修改此 MySQL INSERT 查询？

Question

我正在尝试插入艺术家、歌曲和唱片公司的记录，同时检查数据是否已存在于数据库中。

以下代码来自Mike Fenwick .

    <?php
    $query = "SELECT id FROM table WHERE unique1=value1 AND unique2=value2…";
    $select_result = mysql_query($query);
    if (!mysql_num_rows($select_result)) {
            $query = "INSERT INTO table SET unique1=value1 AND unique2=value2…";
            $insert_result = mysql_query($query);
            $id = mysql_insert_id();
    }
    else {
            $row = mysql_fetch_assoc($select_result);
            $id = $row['id'];
    }
    return $id;
    ?>

我需要修改它以检查是否已经存在三个唯一值(来自 3 个单独的表)，如果不存在，则插入它们。这是我的尝试:

<?php   

            $query = "SELECT id FROM artistsTable WHERE artistName='Beyonce'";
            $select_result = mysql_query($query);
            if (!mysql_num_rows($select_result)) {
                    $query = "INSERT INTO table SET artistName='Beyonce' AND artistImage='beyonce.jpg'";
                    $insert_result = mysql_query($query);
                    $artistID = mysql_insert_id();
            }
            else {
                    $row = mysql_fetch_assoc($select_result);
                    $artistID = $row['artistID'];
            }
            return $artistID;


            $query = "SELECT id FROM recordLabelTable WHERE labelName='Columbia Records'";
                $select_result = mysql_query($query);
                if (!mysql_num_rows($select_result)) {
                    $query = "INSERT INTO table SET labelName='Columbia Records'";
                    $insert_result = mysql_query($query);
                    $labelID = mysql_insert_id();
            }
            else {
                    $row = mysql_fetch_assoc($select_result);
                    $labelID = $row['labelID'];
            }
            return $labelID;



            $query = "SELECT id FROM songTable WHERE trackTitle='Crazy in Love' AND artistID=".$artistID." AND labelID=".$labelID."";
                $select_result = mysql_query($query);
                if (!mysql_num_rows($select_result)) {
                    $query = "INSERT INTO songTable SET trackTitle='Crazy in Love' AND artistID=".$artistID." AND labelID=".$labelID."";
                    $insert_result = mysql_query($query);
                    $songID = mysql_insert_id();
            }
            else {
                    $row = mysql_fetch_assoc($select_result);
                    $songID = $row['songID'];
            }
            return $songID;


    ?>

我假设一定有更有效的方法来做到这一点。任何想法将不胜感激。

Answer 1

使用基本的插入/忽略语法你可以做这样的事情。

插入艺术家详细信息和标签详细信息的几个插入，然后是基于 SELECT 的插入:-

<?php   
    $query = "INSERT IGNORE INTO artistTable (artistName, artistImag) VALUES('Beyonce', 'beyonce.jpg')";
    $insert_result = mysql_query($query);

    $query = "INSERT IGNORE INTO labelTable (labelName) VALUES('Columbia Records')";
    $insert_result = mysql_query($query);

    $query = "INSERT IGNORE INTO songTable (trackTitle, artistID, labelID)
            SELECT 'Crazy in Love', a.artistID, b.labelID
            FROM artistTable a
            INNER JOIN labelTable b
            ON a.artistName = 'Beyonce'
            AND a.artistImag = 'beyonce.jpg'
            AND b.labelName = 'Columbia Records'";
    $insert_result = mysql_query($query);
    $songID = mysql_insert_id();
    return $songID;
?>

正如@LoganWayne 所说，您可能应该使用 MySQLi。