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php - 无法从 mysqli 查询中回显数据

问题描述

我得到了这个查询：

SELECT * 
FROM track 
LEFT JOIN likes ON likes.trackid = track.trackid AND likes.userid='$userid' 
WHERE likes.userid is null ORDER BY rand() LIMIT 1;

现在我无法<?php echo $row['trackid']; ?>从“track”表中回显（但是“likes”表中有一行具有相同名称，我认为这可能是一个问题，但我无法弄清楚）。

我希望你们都能解决我遇到的问题，否则我会尽力解释得更好！

标签： phpsqlmysqli

解决方案

尝试使用显式列名和别名来获取不同 colname 的结果

SELECT likes.trackid like_trackid, track.trackid track_trackid
 FROM track 
LEFT JOIN likes ON likes.trackid = track.trackid AND likes.userid='$userid' 
WHERE likes.userid is null 
ORDER BY rand() LIMIT 1;


<?php echo $row['like_trackid']; ?>

<?php echo $row['track_trackid']; ?>

或者您可以尝试对一个表使用 select *，对第二个数学列使用显式列

SELECT likes.*, track.trackid track_trackid
 FROM track 
LEFT JOIN likes ON likes.trackid = track.trackid AND likes.userid='$userid' 
WHERE likes.userid is null 
ORDER BY rand() LIMIT 1;

<?php echo $row['trackid']; ?>  // this came from  the select *

<?php echo $row['track_trackid']; ?>

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