Question

php将SQL查询结果赋值给变量

$sql = "select field1 from pre_common_member_profile where uid='$username'";
得到的结果是这样的：
field1
1234
怎么把1234赋值给变量

Answer 1

你是只取这一个值还是要取数组。只取一个的话。
$sql = "select field1 from pre_common_member_profile where uid='$username'";
$query = mysql_query($sql);
$bianliang = mysql_result($query,0);
echo $bianliang;
取数组的话。
$sql = "select field1 from pre_common_member_profile where uid='$username'";
$query = mysql_query($sql);
while($row = mysql_fetch_array($query))
{
echo $row["field1 "];
}
完毕。这些操作sql的基础知识，去看看php100的视频教程吧。给分。

Answer 2

$sql = "select field1 from pre_common_member_profile where uid='$username'";
$result = mysql_query($sql);
while($row = mysql_fetch_array($result)){
$number = $row[0]; //按照索引来获取值;然后进行变量的赋值;
}

Answer 3

将"while($row = mysql_fetch_array($result)){" 改为 $row = mysql_fetch_array($result);
因为循环结束时$row的结果会为空.

追问

echo出来还是空的

追答

你的SQL命令是正确的吗?我看到查询条件是:where uid='$username' ,通常uid是id值,而你后面是$username,是不是$id,你写错了?

追问

因为定义变量的时候用的是$username，所以就一直用$username了，要不你直接给个范例，我看下哪里有问题

Answer 4

mysql_query()

mysql_fetch_array

追问

$sql = "select field1 from pre_common_member_profile where uid='$username'";
$result = mysql_query($sql);
while($row = mysql_fetch_array($result)){
$number = $row['filed1'];
echo $number;
输出为空