[英]how to fetch data from one table and insert into another table in php
<?php
$m = "arushi";
$em = "SELECT emailid FROM tblregister WHERE name='$m'";
$q = mysql_query($em);
$n = mysql_fetch_assoc($q);
$fullName = mysql_real_escape_string($_POST['name']);
$address = mysql_real_escape_string($_POST['address']);
$mobNo = mysql_real_escape_string($_POST['dinner']);
$summary = "jsf";
$sql = "INSERT INTO tbljcustomer VALUES('$m', '$n', '$fullName',
'$address','$mobNo', '$summary')";
if(!(mysql_query($sql)))
{
echo "Sorry!!! we were unable to process please try again";
}
else
{
echo "customized";
}
?>
执行此操作一切正常,但它不从tblregister获取emailid,而是显示数组或有时显示资源ID#10。 提前致谢
用这个:
$n=mysql_fetch_assoc($q);
$emailid = $n['emailid'];
并且您的查询将更改为
$sql="insert into tbljcustomer values('$m', '$emailid', '$fullName','$address','$mobNo', '$summary')";
你必须使用: $n['emailid']
$ n是一个关联数组。 查看mysql_fetch_assoc的文档。 使用var_dump($n);
如果你想看到数组的结构。 从文档:
mysql_fetch_assoc返回与获取的行对应的字符串关联数组,如果没有更多行,则返回FALSE。
mysql_fetch_assoc返回一个数组尝试:
insert into tbljcustomer values('$m', '".$n['emailid']."', '$fullName',...
更换
$n
同
$n['emailid']
在您的插入查询中
这将是
$email = $n['emailid'];
$sql="insert into tbljcustomer values('$m', '$email', '$fullName',
'$address','$mobNo', '$summary')";
更换
$sql = "INSERT INTO tbljcustomer VALUES('$m', '$n', '$fullName',
'$address','$mobNo', '$summary')";
同
$sql = "INSERT INTO tbljcustomer VALUES('$m', '".$n['emailid']."', '$fullName',
'$address','$mobNo', '$summary')";
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