編寫PHP代碼中動態由JavaScript

Question

加我有如下代碼，使表：

echo("<script type='text/javascript'> 

var table = document.getElementById('show_member'); 

// Create an empty <tr> element and add it to the 1st position of the table: 
var row = table.insertRow(i); 
i = i+1; 
// Insert new cells (<td> elements) at the 1st and 2nd position of the new <tr> element: 
var cell1 = row.insertCell(0); 
var cell2 = row.insertCell(1); 
var cell3 = row.insertCell(2); 
var cell4 = row.insertCell(3); 
var cell5 = row.insertCell(4); 
var cell6 = row.insertCell(5); 
var cell7 = row.insertCell(6); 
var cell8 = row.insertCell(7); 
var cell9 = row.insertCell(8); 
var cell10 = row.insertCell(9); 
var cell11 = row.insertCell(10); 

// Add some text to the new cells: 
cell1.innerHTML = '$name'; 
cell2.innerHTML = '$name2'; 
cell3.innerHTML = '$lastname'; 
cell4.innerHTML = '$lastname2'; 
cell5.innerHTML = '$mellicode'; 
cell6.innerHTML = '$estekhdamnum'; 
cell7.innerHTML = '$email'; 
cell8.innerHTML = '$username'; 
cell9.innerHTML = '$password'; 
cell10.innerHTML = '$id'; 
var checkbox = document.createElement('INPUT'); 
***checkbox.type = 'checkbox'; 
checkbox.name = 'mark[]'; 
checkbox.value = i-1; 
cell11.appendChild(checkbox);*** 
</script>"); 

和下方的按鈕：

<div class="container"> 
    <input type="button" class="form-control" name = "karbar" id ="check" value="افزودن به لیست کاربران"onclick="add_karbar1()"/><br><br> 
</div> 

以下是我的PHP代碼更新表格

<?php 
if(isset($_POST['karbar'])){ 
    //echo "<script>alert('was selected')</script>"; 
    foreach($_REQUEST['mark'] as $value){ 
     echo "<script>alert('$value was selected');</script>"; 
    } 
} 



?> 

它給了我這個錯誤，爲什麼？

說明：未定義指數：標記在E：\ Xampp_installed \ htdocs中\上線路卡\ add_karbar.php 236

警告：在E中的foreach（）提供參數無效：\ Xampp_installed \ htdocs中\卡\ add_karbar.php on line 23

如何識別選中的複選框併爲每行中的每個選中的複選框執行php代碼？

感謝

Answer 1

你的問題可能造成未允許腳本到達checkbox.name = 'mark[]';線***一個JavaScript語法錯誤。如果該行不執行，則$_REQUEST中將不會有mark變量。更改

***checkbox.type = 'checkbox'; 
checkbox.name = 'mark[]'; 
checkbox.value = i-1; 
cell11.appendChild(checkbox);*** 

到

checkbox.type = 'checkbox'; 
checkbox.name = 'mark[]'; 
checkbox.value = i-1; 
cell11.appendChild(checkbox); 

，看看它是否工作