我想从一个 .csv 文件中导出没有连接的 3 个表中的数据。 我正在尝试连接，但没有得到我想要的结果。

下面是我的表结构

Playlist Songs Rating

代码

$mysql_host = DB_HOST;
$mysql_user = DB_USER;
$mysql_pass = DB_PASSWORD;
$mysql_db = DB_NAME;

$pre = $wpdb->prefix;

$link = mysql_connect($mysql_host, $mysql_user, $mysql_pass) or die('Could not connect: ' . mysql_error());

mysql_select_db($mysql_db, $link) or die('Could not select database: ' . $mysql_db);

$query = "SELECT plist.*, psong.*, prate.* 
          FROM " . $pre . "foo_playlists As plist 
          LEFT JOIN " . $pre . "foo_songs As psong
          On plist.playlist_name = psong.splaylist_name 
          LEFT JOIN " . $pre . "foo_rating As prate
          On psong.song_id = prate.rsong_id";

$result = mysql_query($query);
$row = mysql_fetch_assoc($result);

$line = "";
$comma = "";

foreach ($row as $name => $value) {
   $line .= $comma . '"' . str_replace('"', '""', $name) . '"';
   $comma = ",";
}
    $line .= "\n";
    $out = $line;
    mysql_data_seek($result, 0);
    while ($row = mysql_fetch_assoc($result)) {
        $line = "";
        $comma = "";
        foreach ($row as $value) {
            $line .= $comma . '"' . str_replace('"', '""', $value) . '"';
            $comma = ",";
        }
        $line .= "\n";
        $out.=$line;
    }
    $csv_file_name = 'songs_' . date('Ymd_His') . '.csv'; # CSV FILE NAME WILL BE table_name_yyyymmdd_hhmmss.csv
    header("Content-type: text/csv");
    header("Content-Disposition: attachment; filename=" . $csv_file_name);
    header("Content-Description:File Transfer");
    header('Content-Transfer-Encoding: binary');
    header('Cache-Control: must-revalidate, post-check=0, pre-check=0');
    header('Pragma: public');
    header('Content-Type: application/octet-stream');
    echo __($out, "foo");

    exit;

我得到了 this result我想要这个 desired result

我该怎么做？

最佳答案

那么，您的问题是您无法在一个唯一的 MySQL 查询中同时检索所有数据，因为它们不是相关数据。您的问题只是输出，因此，您只需将 3 组结果关联到一个数组中。为此:

• 执行三个查询，分别保存在三个不相关的数组中。

• 将它们与您将与所有人共享的 key 联系起来。

• 遍历所有数组，将值分配给一个主要的“输出”数组。

有了它，您将拥有可以输出的数组以获取所需的 CSV。为了这个例子，由于我不能用你的变量和查询写一个有效的代码，我写了下面的例子。它有 3 个不同的数组，你必须从你的数据库中获取模拟数据，但你可以捕获这个想法。只需复制并粘贴，您就会获得实际示例:

<?php

$playlists  = array(
    array(
        'id'    => 1
    ,   'data'  => 'playlist1'
    )
,   array(
        'id'    => 2
    ,   'data'  => 'playlist2'
    )
,   array(
        'id'    => 3
    ,   'data'  => 'playlist3'
    )
);

$songs  = array(
    array(
        'id'    => 1
    ,   'data'  => 'song1'
    )
,   array(
        'id'    => 2
    ,   'data'  => 'song2'
    )
,   array(
        'id'    => 3
    ,   'data'  => 'song3'
    )
,   array(
        'id'    => 4
    ,   'data'  => 'song4'
    )
,   array(
        'id'    => 5
    ,   'data'  => 'song5'
    )
);
$rates  = array(
    array(
        'id'    => 1
    ,   'data'  => 'rating1'
    )
,   array(
        'id'    => 2
    ,   'data'  => 'rating2'
    )
,   array(
        'id'    => 3
    ,   'data'  => 'rating3'
    )
,   array(
        'id'    => 4
    ,   'data'  => 'rating4'
    )
,   array(
        'id'    => 5
    ,   'data'  => 'rating5'
    )
,   array(
        'id'    => 6
    ,   'data'  => 'rating6'
    )
);

// Count all the arrays and get the bigger:
$num        = 0;
$play_num   = count( $playlists );
$num        = ($play_num > $num) ? $play_num : $num;

$song_num   = count( $songs );
$num        = ($song_num > $num) ? $song_num : $num;

$rate_num   = count( $rates );
$num        = ($rate_num > $num) ? $rate_num : $num;


$output = array();
for ( $i = 0; $i<=$num; $i++ ) {
    $output[]   = array(
        'id_playlist'   => !empty( $playlists[$i]['id'] )   ? $playlists[$i]['id']   : ''
    ,   'data_playlist' => !empty( $playlists[$i]['data'] ) ? $playlists[$i]['data'] : ''
    ,   'id_song'       => !empty( $songs[$i]['id'] )       ? $songs[$i]['id']       : ''
    ,   'data_song'     => !empty( $songs[$i]['data'] )     ? $songs[$i]['data']     : ''
    ,   'id_rate'       => !empty( $rates[$i]['id'] )       ? $rates[$i]['id']       : ''
    ,   'data_rate'     => !empty( $rates[$i]['data'] )     ? $rates[$i]['data']     : ''
    );
}

foreach ( $output as $out ) {
    echo implode( ' - ', $out);
    echo '<br>';
}

输出:

1 - playlist1 - 1 - song1 - 1 - rating1
2 - playlist2 - 2 - song2 - 2 - rating2
3 - playlist3 - 3 - song3 - 3 - rating3
  -           - 4 - song4 - 4 - rating4
  -           - 5 - song5 - 5 - rating5
  -           -   -       - 6 - rating6

关于php - 如何从一个 .csv 文件 php 中的 3 个表导出数据，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/25116730/