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like 查询,报错,求指点

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Release: 2016-06-23 14:13:40
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      mysql_connect("localhost", "root", "123") or die("Could not connect: " . mysql_error());	 mysql_select_db("iweibo");    $result = mysql_query("SELECT * FROM user where username like %admin%");    while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {			var_dump($row);    }
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报错信息:

Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\34\test.php on line 19


回复讨论(解决方案)

$result = mysql_query("SELECT * FROM user where username like '%admin%'");

肯定是SQL语句的问题,echo 你的sql语句,拿到数据库去运行就知道什么原因了。
$result = mysql_query("SELECT * FROM user where username like '%admin%'");
试一试。

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