<?php
$title=$_POST[search];
$title= "% ".$title. "% ";
$query = "select * from mytable where title like '$title' ";
$result = mysql_query($query);
$row =mysql_fetch_array($result);
while($row){
echo $row['title'];
}
?>代码如上,不知道哪错了!!!总是Warning: mysql_fetch_row(): supplied argument is not a valid MySQL result resource in D:\Www_local\yy\file.php on line 14,,,,求帮忙
$title=$_POST[search];
$title= "% ".$title. "% ";
$query = "select * from mytable where title like '$title' ";
$result = mysql_query($query);
$row =mysql_fetch_array($result);
while($row){
echo $row['title'];
}
?>代码如上,不知道哪错了!!!总是Warning: mysql_fetch_row(): supplied argument is not a valid MySQL result resource in D:\Www_local\yy\file.php on line 14,,,,求帮忙
<?php
$title=$_POST["search"]; //加个引号,不管单引还是双的
$title= "'%".$title."%'";
$query = "select * from mytable where title like {$title}";
$result = mysql_query($query);
while($row =mysql_fetch_array($result)){
echo $row['title'];
}
?>
echo $query;
得到的结果:select * from mytable where title like '% ddd% ' 建议:$title= "% ".$title. "% ";
修改为:$title= "%".$title."%";代码应该是正常的。
去掉空格试一试
Warning: mysql_fetch_row() 怎么提示这个方法有错了??
看你代码没有这个函数吧