php删除问题

我要删除一条数据

<?php
$cha=$_POST["cha"];

$id = mysql_connect("localhost","root","root");
mysql_select_db("login", $id);
$result = mysql_query ("SELECT name,s_name,context,pic_name FROM
u_gg",$id);
print "<table border='1'>";
print "<th align = 'center'>广告名字</th>";
print "<th align = 'center'>商家姓名</th>";
print "<th align = 'center'>广告内容</th>";
print "<th align = 'center'>广告图片</th>";

while ($row = mysql_fetch_array($result))
{
print "<tr> \n";
print "<td align = 'center'>{$row["name"]}</td>";
print "<td align = 'center'>{$row["s_name"]}</td>";
print "<td align = 'center'>{$row["context"]}</td>";
print "<td align = 'center'> <img src=".$row["pic_name"]."</td> <br> </n>";
print "</tr> \n";
}
$chaxun=mysql_query("DELETe * FROM u_gg WHERe name='$cha'",$id) ;
print "</table> \n";
?>

但是删除不了记录的?那里出错了?

print "<td align = 'center'>{$row["name"]}</td>";
print "<td align = 'center'>{$row["s_name"]}</td>";
print "<td align = 'center'>{$row["context"]}</td>";

这里出错了，你看下双引号和单引号，$row里的两个双引号没有自身配对，而是与外层双引号配对了

while ($row = mysql_fetch_array($result)) { .......

}

这句就是在$row为真的时候执行{}里的语句,既然

$chaxun=mysql_query("DELETe * FROM u_gg WHERe name=$row['name']",$id) ;

已经在{}之后了,那说明$row已经为假,这样$row['name']是没有值的,

你在{}之后打印一下$row['name']就知道了.

还有,在php里输出用echo ,关于echo和print的区别,到网上搜索一下大把.

把这句

$chaxun=mysql_query("DELETe * FROM u_gg WHERe name='$cha'",$id) ;

改为

$chaxun=mysql_query("DELETe * FROM u_gg WHERe name=$row['name']",$id) ;

即可。