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php - LIKE 运算符和 NOT IN 的正确用法

转载 作者:行者123 更新时间:2023-11-29 05:35:17 25 4
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Possible Duplicate:
mysql_fetch_array() expects parameter 1 to be resource, boolean given in select

这目前有效...

$resultInput = mysql_query("SHOW COLUMNS FROM " . $table. " WHERE Field NOT IN ('id', 'created', 'date_modified', 'content', 'type', 'bodytext', 'project_content', 'content_short') AND Field NOT LIKE '%_image%'");

不过,我想删除所有名称为 content 的字段,并将其添加到类似 %content%

的 LIKE 函数中
$resultInput = mysql_query("SHOW COLUMNS FROM " . $table. " WHERE Field NOT IN ('id', 'created', 'date_modified', 'type', 'bodytext') AND Field NOT LIKE ('%_image%', %content%)");

但这似乎不起作用?并返回一个“

Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given in
C:\xampp\htdocs\framework.php on line 33

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