我有一个php文件,它连接到一个MySQL数据库。它应该从URL中获得两个参数,并对参数string使用一个like函数。这是我的php文件:
<?php
mysql_connect("mysql.1freehosting.com","u948577195_uname","p7CraCuRAw");
mysql_select_db("u948577195_dbnam");
$param = $_GET['param'];
$UserID = $_GET['UserID'];
$sql=mysql_query("SELECT UserID, UserName, DisplayName
FROM User
WHERE (UserID <> '$UserID') and (UserID NOT IN
(SELECT UserID1 AS UserID
FROM Friend
WHERE (UserID2 = '$UserID')
UNION ALL
SELECT UserID2 AS UserID
FROM Friend AS Friend_1
WHERE (UserID1 = '$UserID'))) AND( (UserName LIKE %'$param'%) OR
(DisplayName LIKE %'$param'%))");
while($row=mysql_fetch_assoc($sql))
$output[]=$row;
print(json_encode($output));
mysql_close();
?>这是php文件的URL:project/query7.php?param=s&UserID=1。这个功能似乎是因为类似的部分而下降的。我该怎么解决呢?
发布于 2014-07-13 18:35:37
您的查询格式如下所示。将%包含在''中
$sql=mysql_query("SELECT UserID,
UserName,
DisplayName
FROM User
WHERE UserID <> '$UserID'
and UserID NOT IN (
SELECT UserID1 AS UserID
FROM Friend
WHERE UserID2 = '$UserID'
UNION ALL
SELECT UserID2
FROM Friend
WHERE UserID1 = '$UserID'
)
AND (UserName LIKE '%$param%'
OR DisplayName LIKE '%$param%')")https://stackoverflow.com/questions/24725894
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