Question

我写了一个MySQL查询：

$feedItem->tags =

$result = mysqli_query("SELECT * 
    FROM qzxh_k2_tags, qzxh_k2_tags_xref
    WHERE qzxh_k2_tags.id = qzxh_k2_tags_xref.tagID
    AND qzxh_k2_tags.id = '406'
    AND qzxh_k2_tags_xref.itemID = '".$item->id"'");

while($tag = mysqli_fetch_array($result))
echo $tag;

查询本身在PHPMyAdmin中显示了我需要的结果，但由于某种原因，在PHP中使用时没有显示任何内容。有什么想法，我出错了吗？

Answer 1

您缺少连接运算符（点）

更改

"SELECT * 
    FROM qzxh_k2_tags, qzxh_k2_tags_xref
    WHERE qzxh_k2_tags.id = qzxh_k2_tags_xref.tagID
    AND qzxh_k2_tags.id = '406'
    AND qzxh_k2_tags_xref.itemID = '".$item->id"'"
// You are missing dot . here                  ^

到

"SELECT * 
FROM qzxh_k2_tags, qzxh_k2_tags_xref
WHERE qzxh_k2_tags.id = qzxh_k2_tags_xref.tagID
AND qzxh_k2_tags.id = '406'
AND qzxh_k2_tags_xref.itemID = '".$item->id."'"

Answer 2

在sql查询中使用正确的字符串连接。查找以下更正的查询。

$result = mysqli_query("SELECT * 
    FROM qzxh_k2_tags, qzxh_k2_tags_xref
    WHERE qzxh_k2_tags.id = qzxh_k2_tags_xref.tagID
    AND qzxh_k2_tags.id = '406'
    AND qzxh_k2_tags_xref.itemID = '".$item->id."'");

MySQL查询在PHPMyAdmin中工作但不能在PHP中工作

2 个答案: