Question

我正在使用AJAX将隐藏的输入值传递给另一页并显示在其中，但价值没有传递。我是PHP，jQuery和AJAX的新手。有人可以帮我吗？

index.php：

<?php 
  foreach($imgs as $keys => $vals)
  {
    echo '<img class="myimg1" id="myimg" src="' . "images/$vals" . '" width="20%" height="20%"/>';
    echo '<input type="hidden" value="'.$vals.'" name="delete_file" id="delete_file" />';
    echo '<input type="button" value="Delete image" onclick="delete_image()"/>';
  }
?>

js

function delete_image() {
  var status = confirm("Are you sure you want to delete ?");  
  if (status == true) {
    var file = $("#delete_file").val();
    $.ajax({
      type: "POST",
      url: "action.php",
      data: { file: file },
      success(html) {
        alert('Deleted');
      }
    });
  }
}

action.php

<?php
  $num = $_POST["file"];
  echo $num;
?>

Answer 1

我希望以下解决方案能够为您提供帮助。

我将输入类型按钮中的onclick函数和功能修改为delete_name（ele）

var file = ele;

echo '<img class="myimg1" id="myimg" src="' . "images/$vals" . '" width="20%" height="20%"/>';
    echo '<input type="hidden" value="'.$vals.'" name="delete_file" id="delete_file" />';
    echo '<input type="button" value="Delete image" onclick="delete_image('.$vals.')"/>';

function delete_image(ele) {
  var status = confirm("Are you sure you want to delete ?");  
  if (status == true) {
    var file = ele;
    $.ajax({
      type: "POST",
      url: "action.php",
      data: { file: file },
      success(html) {
        alert('Deleted');
      }
    });
  }
}

Answer 2

您正在循环播放图像，所有这些图像都具有相同的名称和ID。不好我将对其进行更改，以便他们都获得不同的名称和ID。

<?php 
  $counter = 0;
  foreach($imgs as $keys => $vals)
  {
    echo '<img class="myimg1" id="myimg" src="' . "images/$vals" . '" width="20%" height="20%"/>';
    echo '<input type="hidden" value="'.$vals.'" name="delete_file[' . $counter . ']" id="delete_file_' . $counter . '" />';
    echo '<input type="button" value="Delete image" onclick="delete_image()"/>';
    $counter++;
  }
?>

您还可以将图像放入表单中，并使用jQuery序列化以通过Ajax发送。

还更改以下内容的成功部分以查看PHP错误：

function delete_image() {
  var status = confirm("Are you sure you want to delete ?");  
  if (status == true) {
    var file = $("#delete_file").val();
    $.ajax({
      type: "POST",
      url: "action.php",
      data: { file: file },
      success: function(data, textStatus, jqXHR) {
        // use console.log to find PHP errors in the console of the browser
        console.log(data); 
        // you can alert data or use is to check for errors of whatever you want
        if (data == 'success') { alert(data); } 
      }
    });
  }
}

值是否没有使用AJAX传递到另一页？

2 个答案: