【发布时间】：2014-08-16 08:07:51
【问题描述】：
我有一个 php 生成的表格/表格，带有这样的复选框：
if ($query) {
if (!mysqli_num_rows($query)) {
//Empty storage
echo "There are no items in '$storage'.";
exit();
} else {
//form
echo "<form name='send_parts_form' action='../includes/generatetab.php' method='post'>";
//Table header
echo "
<table>
<tr>
<th>Part</th>
<th>PN</th>
<th>Manufactured</th>
<th>Serial</th>
<th>Site</th>
<th>Date replaced</th>
<th>By user</th>
<th>Faulty</th>
<th>Send</th>
<th>Select</th>
</tr>";
//Retrieved data
while ($row = mysqli_fetch_array($query)) {
echo "<tr>";
echo "<td>" . $row['part_type'] . "</td>";
echo "<td>" . $row['pn'] . "</td>";
echo "<td>" . $row['manufactured'] . "</td>";
echo "<td>" . $row['serial'] . "</td>";
echo "<td>" . $row['site_id'] . "</td>";
echo "<td>" . $row['date'] . "</td>";
echo "<td>" . $row['user'] . "</td>";
echo "<td>" . $row['faulty'] . "</td>";
echo "<td><input type='checkbox' name='send_parts[]' class='checkclass' value=" . $row['id'] . "></td>";
echo "</tr>";};
echo "</table>";
echo "</br>";
echo "</br>";
echo "<input type='button' onclick='sendToZG()' value='Send'/>";
echo "</br>";
echo "<input type='submit' name='submit' value='Generate tab' />";
echo "</form>";
exit();
}
} else {
die("Query failed");
}
然后，用户检查他们想要的选项，并在提交（生成选项卡）时获得选项卡分隔的文本以及他们选择的值。 我现在希望当他们单击“发送”时将值发布到另一个 php 页面并在同一页面上返回结果（在 SentList div 下）。我有这样的js：
//Browser Support Code
function sendToZG(){
var ajaxRequest; // The variable that makes Ajax possible!
try{
// Opera 8.0+, Firefox, Safari
ajaxRequest = new XMLHttpRequest();
}catch (e){
// Internet Explorer Browsers
try{
ajaxRequest = new ActiveXObject("Msxml2.XMLHTTP");
}catch (e) {
try{
ajaxRequest = new ActiveXObject("Microsoft.XMLHTTP");
}catch (e){
// Something went wrong
alert("Your browser broke!");
return false;
}
}
}
ajaxRequest.onreadystatechange = function(){
if(ajaxRequest.readyState == 4 && ajaxRequest.status == 200){
var ajaxDisplay = document.getElementById('SentList');
ajaxDisplay.innerHTML = ajaxRequest.responseText;
}
}
// Now get the value from page and pass it to server script.
var formData = new FormData( document.getElementsByName("send_parts")[0] );
ajaxRequest.open('POST', "../includes/sendtozg.php", true);
ajaxRequest.setRequestHeader("Content-Type", "application/x-www-form-urlencoded");
ajaxRequest.send("send_parts=" + formData);
}
编辑：ajaxRequest.send("send_part=" + formData);到 ajaxRequest.send("send_parts=" + formData); 现在它返回：
在第 53 行为 foreach() 提供的参数无效（这是我在 sendtozg.php 中获取数据的地方）。
我会在文章末尾添加 sendtozg.php。
如果不是：
<form name='send_parts_form' action='../includes/generatetab.php' method='post'>
我回应：
<form name='send_parts_form' action='../includes/sendtozg.php' method='post'>
提交后，脚本 sendtozg.php 可以正常执行，但在不同的页面上。
所以基本上我要做的是为 php 生成的表单提供 2 个选项：
- 生成制表符分隔的 txt 文件
- 执行 sendtozg.php 并在同一页面返回结果
我已经有了两个脚本（generatetab.php 和 sendtozg.php），它们工作正常。
sendtozg.php:
if (!empty($_POST['send_parts'])){
foreach ($_POST['send_parts'] as $send_parts){
$getchecked = mysqli_query($con, "SELECT * FROM $storage WHERE id=$send_parts");
while ($row = mysqli_fetch_array($getchecked)) {
$copypart = mysqli_query($con, "INSERT INTO sent_parts (part_type, pn, manufactured, serial, site_id, date, user, faulty, log)
SELECT part_type, pn, manufactured, serial, site_id, date, user, faulty, log
FROM $storage WHERE id=$send_parts");
// check to see if it copied
$getserial = mysqli_query($con, "SELECT serial FROM $storage WHERE id=$send_parts");
$getserial_row = mysqli_fetch_array($getserial, MYSQLI_NUM);
$foundserial = $getserial_row[0];
$checkcopy = mysqli_query($con, "SELECT id FROM sent_parts WHERE serial = '$foundserial'");
// add user info and date
$addinfo = mysqli_query($con, "UPDATE sent_parts SET sent2zg='$user', date2zg='$strdate' WHERE serial = '$foundserial'");
if (!$check1_res) {
printf("Error: %s\n", mysqli_error($con));
exit();
};
//delete from storage
if($checkcopy > 0) {
$getpart = mysqli_query($con, "SELECT part_type FROM sent_parts WHERE serial='$foundserial'");
$getpart_row = mysqli_fetch_array($getpart, MYSQLI_NUM);
$deletedpart = $getpart_row[0];
$removepart = mysqli_query($con, "DELETE FROM $storage WHERE id = '$send_parts'");
echo "Part " . $deletedpart . " has been transfered";
} else {
echo "Part " . $row['part_type'] . "was NOT transfered";
};
};
} exit ();
} else {
echo "Nothing was selected, please try again!";
}
【问题讨论】：
-
您是否尝试过制作第三个 PHP 文件，仅使用
include '../includes/generatetab.php';和include '../includes/sendtozg.php';，然后将表单发送到该文件？ -
不会同时执行这两个脚本吗？我想把它们分开，所以可以单独执行。
-
啊，我明白了，根据用户操作执行
generatetab.php或sendtozg.php。对吗？ -
是的，没错！它们都可以通过 xmlhttprequest 或其他方式执行。重要的是sendtozg.php在同一页面上返回结果，generatetab.php无论如何都会下载txt文件。
标签： php forms xmlhttprequest