[英]PHP POST form after AJAX call
問題:我的 php 表單沒有提交。
這是我的頁面:
它有一個 php 渲染的顏色列表:
<div class="table-container">
<div>
<table class="myTable" id="myTable">
<tr class="header">
<th>Variants</th>
<th>Size</th>
<th>Price (€)</th>
</tr>
<?php
$current_name = $_GET['prod-name'];
$get_product_det = "SELECT * FROM product_details WHERE product_name='$current_name' ORDER BY position";
$run_product_det = mysqli_query($con, $get_product_det);
while ($row_product_det = mysqli_fetch_array($run_product_det)){
$id = $row_product_det['id'];
$product_variant = $row_product_det['product_variant'];
$product_size = $row_product_det['size'];
$product_price = $row_product_det['price'];
$position = $row_product_det['position'];
echo "<tr data-index='$id' data-position='$position'>
<td><a id='$id'>$product_variant</a></td>
<td>$product_size</td>
<td>$product_price</td>
</tr>";
};
?>
</table>
</div>
每種顏色都有唯一的 ID。 單擊顏色會觸發工作正常的 AJAX 腳本:
<div class='product-det-div' id='product_details'>
<script>
var links = document.getElementsByTagName('a');
for (var i = 0, il = links.length; i < il; i++) {
links[i].onclick = function() {
var id = this.id;
var product_details = document.getElementById('product_details');
var request = new XMLHttpRequest();
request.open('POST', 'product_details.php?variant_id=' + id, true);
request.onreadystatechange = function() {
if (request.readyState === 4 & request.status === 200) {
product_details.innerHTML = request.responseText;
} else {
product_details.innerHTML = 'An error occurred during your request: ' + request.status + ' ' + request.statusText;
}
};
request.send();
};
};
</script>
</div>
當ajax調用發生時,頁面的URL會發生什么? 我正在通過 URL 傳遞顏色的 ID,但是當嘗試使用 PHP 獲取它時,它似乎沒有找到它。 這是頁面的代碼:
$current_id = $_REQUEST['variant_id'];
$get_product_det = "SELECT * FROM product_details WHERE id=$current_id";
$run_product_det = mysqli_query($con, $get_product_det);
while ($row_product_det = mysqli_fetch_array($run_product_det)){
$product_variant = $row_product_det['product_variant'];
$product_size = $row_product_det['size'];
$product_price = $row_product_det['price'];
echo "
<form action='' method='post'>
<h2 style='margin-bottom: 20px;'>$product_variant</h2>
<div><label>Nome Prodotto</label><input value='$product_variant'></div>
<div><label>Dimensione</label><input value='$product_size' type='number' name='product_size'></div>
<div><label>Prezzo (€)</label><input value='$product_price' id='product_price' name='product_price'></div>
<button type='submit' name='edit_variant_btn'>Send</button>
</form>
";
if(isset($_POST['edit_variant_btn'])) {
$variant = $_POST['product_size'];
$current_id = $_REQUEST['variant_id'];
$update_size = "UPDATE product_details SET size = '$variant' WHERE id = '$current_id'";
$run_update = mysqli_query($con, $update_size);
if($run_update) {
echo "<script>window.open('variable_product.php?prod-name=Polycolor', '_self');</script>";
}
}
};
?>
感謝您的時間,任何幫助表示贊賞。
編輯:我嘗試了你們建議的所有更改,也嘗試了 $_REQUEST["variant_id"] 正如 Banujan Balendrakumar 所說,但仍然沒有結果。 我發現使其工作的唯一方法是從這里更改表單操作:
<form action='' method='post'>
到
<form action='product_details.php?variant_id=$current_id' method='post'>
這樣做是因為在單擊按鈕時,它會打開該頁面並從那里獲取 ID 值,但這只是解決問題的一種方法......還有其他想法嗎?
將$ _GET [“ variant_id”]替換為$ _REQUEST [“ variant_id”]
對我有用...
<div class="table-container">
<div>
<table class="myTable" id="myTable">
<tr class="header">
<th>Variants</th>
<th>Size</th>
<th>Price (€)</th>
</tr>
<?php
$con = new mysqli('localhost','root','','check');
$get_product_det = "SELECT * FROM product_details ORDER BY position";
$run_product_det = mysqli_query($con, $get_product_det);
while ($row_product_det = mysqli_fetch_array($run_product_det)){
$id = $row_product_det['id'];
$product_variant = $row_product_det['product_variant'];
$product_size = $row_product_det['size'];
$product_price = $row_product_det['price'];
$position = $row_product_det['position'];
echo "<tr data-index='$id' data-position='$position'>
<td><a id='$id'>$product_variant</a></td>
<td>$product_size</td>
<td>$product_price</td>
</tr>";
};
?>
</table>
</div>
<div class='product-det-div' id='product_details'>
<script>
var links = document.getElementsByTagName('a');
for (var i = 0, il = links.length; i < il; i++) {
links[i].onclick = function() {
var id = this.id;
var product_details = document.getElementById('product_details');
var request = new XMLHttpRequest();
request.open('POST', 'product_details.php?variant_id=' + id, true);
request.onreadystatechange = function() {
if (request.readyState === 4 & request.status === 200) {
product_details.innerHTML = request.responseText;
} else {
product_details.innerHTML = 'An error occurred during your request: ' + request.status + ' ' + request.statusText;
}
};
request.send();
};
};
</script>
</div>
產品詳情
<?php
$current_id = $_REQUEST['variant_id'];
$con = new mysqli('localhost','root','','check');
$get_product_det = "SELECT * FROM product_details WHERE id=$current_id";
$run_product_det = mysqli_query($con, $get_product_det);
while ($row_product_det = mysqli_fetch_array($run_product_det)){
$product_variant = $row_product_det['product_variant'];
$product_size = $row_product_det['size'];
$product_price = $row_product_det['price'];
echo "
<form action='' method='post'>
<h2 style='margin-bottom: 20px;'>$product_variant</h2>
<div><label>Nome Prodotto</label><input value='$product_variant'></div>
<div><label>Dimensione</label><input value='$product_size' type='number' name='product_size'></div>
<div><label>Prezzo (€)</label><input value='$product_price' id='product_price' name='product_price'></div>
<button type='submit' name='edit_variant_btn'>Send</button>
</form>
";
if(isset($_POST['edit_variant_btn'])) {
$variant = $_POST['product_size'];
$current_id = $_REQUEST['variant_id'];
$update_size = "UPDATE product_details SET size = '$variant' WHERE id = '$current_id'";
$run_update = mysqli_query($con, $update_size);
if($run_update) {
echo "<script>window.open('variable_product.php?prod-name=Polycolor', '_self');</script>";
}
}
};
?>
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