无法从 mysqli 查询中回显数据答案

【问题标题】：Can't echo data from mysqli query无法从 mysqli 查询中回显数据
【发布时间】：2021-05-30 11:22:38
【问题描述】：

我收到了这个查询：

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;

现在我无法从“track”表中回显<?php echo $row['trackid']; ?>（但是“likes”表中有一行同名，我认为这可能是个问题，但我想不通出）。

我希望你们都明白我遇到的问题，否则我会尽力解释得更好！

【问题讨论】：

标签： php sql mysqli

【解决方案1】：

尝试使用明确的列名和别名以不同的列名获取结果

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 * 并为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']; ?>

【讨论】：

谢谢，这几乎解决了问题，正如您在我的问题中看到的那样，我希望能够从行中回显所有内容 (*)，这可能吗？