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我试图用这个脚本做的是允许用户更新他们网站的 url，因为每个用户不会拥有相同数量的网站，所以我很难为每个用户添加$_POST['website']。

这是脚本

<?php
include("config.php");
include("header.php");
include("functions.php");

if(!isset($_SESSION['username']) && !isset($_SESSION['password'])){
    header("Location: pubs.php");
    }

$getmember = mysql_query("SELECT * FROM `publishers` WHERE username = '".$_SESSION['username']."'");
$info = mysql_fetch_array($getmember);
$getsites = mysql_query("SELECT * FROM `websites` WHERE publisher = '".$info['username']."'");

$postback = $_POST['website'];
$webname = $_POST['webid'];
if($_POST['submit']){
foreach ( $_POST['website'] as $key => $value )
{
 $update = mysql_query("UPDATE `websites` SET `postback` = '".mysql_real_escape_string($postback[$value])."' WHERE id = '$webname'");
}
}


print"
<div id='center'>
<span id='tools_lander'><a href='export.php'>Export Campaigns</a></span>
<div id='calendar_holder'>
<h3>Please define a postback for each of your websites below. The following variables should be used when creating your postback.<br />
cid = Campaign ID<br />
sid = Sub ID<br />
rate = Campaign Rate<br />
status = Status of Lead. 1 means payable 2 mean reversed<br />
A sample postback URL would be <br />
http://www.example.com/postback.php?cid=#cid&sid=#sid&rate=#rate&status=#status</h3>
<table class='balances' align='center'>
<form method='POST' action=''>";

while($website = mysql_fetch_array($getsites)){
print"
<tr>
<input type ='hidden' name='webid' value='".$website['id']."' />
<td style='font-weight:bold;'>".$website['name']."'s Postback:</td>
<td><input type='text' style='width:400px;' name='website[]' value='".$website['postback']."' /></td>
</tr>";
}
print"
<td style='float:right;position:relative;left:150px;'><input type='submit' name='submit' style='font-size:15px;height:30px;width:100px;' value='Submit' /></td>
</form>
</table>
</div>";

include("footer.php");
?>

我正在尝试做的是将文本框中输入的内容插入到其相应的网站，我想不出任何其他方式来做到这一点，这显然不起作用并返回一条通知，说明Array to string conversion 如果有更合乎逻辑的方式要做到这一点，请告诉我。

更新 我添加了一个 foreach 语句，但这似乎仍然不能解决问题。它不会更新数据库中的任何内容。

4

1 回答 1

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我能够通过一些试验和错误来解决问题，Lawrence 帮助通知我使用 foreach 语句。这就是我的结局。

<?php
include("config.php");
include("header.php");
include("functions.php");

if(!isset($_SESSION['username']) && !isset($_SESSION['password'])){
    header("Location: pubs.php");
    }

$getmember = mysql_query("SELECT * FROM `publishers` WHERE username = '".$_SESSION['username']."'");
$info = mysql_fetch_array($getmember);
$getsites = mysql_query("SELECT * FROM `websites` WHERE publisher = '".$info['username']."'");

$postback = $_POST['website'];
$webname = $_POST['webid'];
if($_POST['submit']){
$i = -1;
foreach ($postback as $key => $value)
{
$i ++;
print_r($webname[$i]);
 $update = mysql_query("UPDATE `websites` SET `postback` = '".cleanQuery($postback[$key])."' WHERE `id` = '".$webname[$i]."'") or die("MySQL ERROR: ".mysql_error());
}
}



print"
<div id='center'>
<span id='tools_lander'><a href='export.php'>Export Campaigns</a></span>
<div id='calendar_holder'>
<h3>Please define a postback for each of your websites below. The following variables should be used when creating your postback.<br />
cid = Campaign ID<br />
sid = Sub ID<br />
rate = Campaign Rate<br />
status = Status of Lead. 1 means payable 2 mean reversed<br />
A sample postback URL would be <br />
http://www.example.com/postback.php?cid=#cid&sid=#sid&rate=#rate&status=#status</h3>
<table class='balances' align='center'>
<form method='POST' action=''>";

while($website = mysql_fetch_array($getsites)){
print"
<tr>
<input type ='hidden' name='webid[]' value='".$website['id']."' />
<td style='font-weight:bold;'>".$website['name']."'s Postback:</td>
<td><input type='text' style='width:400px;' name='website[]' value='".$website['postback']."' /></td>
</tr>";
}
print"
<td style='float:right;position:relative;left:150px;'><input type='submit' name='submit' style='font-size:15px;height:30px;width:100px;' value='Submit' /></td>
</form>
</table>
</div>";

include("footer.php");
?>
于 2012-11-04T01:22:21.127 回答