Javascript 无法在php中从ajax访问2个或更多文件

这是我的PHP代码

for($x=0; $x<count($_FILES['image1']['name']); $x++ ) {

        // echo "Hello";
        $file_name = $_FILES['image1']['name'][$x];

        $file_size = $_FILES['image1']['size'][$x];
        $file_tmp  = $_FILES['image1']['tmp_name'][$x];
        $lclLocation1 = "http://$_SERVER[HTTP_HOST]/images/".$_FILES["image1"]["name"][0];
        $lclLocation2 = "http://$_SERVER[HTTP_HOST]/images/".$_FILES["image1"]["name"][1];


        $t = explode(".", $file_name);
        $t1 = end($t);
        $file_ext = strtolower(end($t));

        $ext_boleh = array("jpg", "jpeg", "png", "gif", "bmp");

        if(in_array($file_ext, $ext_boleh)) {
            $sumber = $file_tmp;
            $tujuan = "../images/" . $file_name;
            // echo "Image got";
            move_uploaded_file($sumber, $tujuan);

        }
        else  {
            echo "Only Images can be store!";
        }
    } // end for

在此之后，我们将使用ajax发送到PHP

$.ajax({
        url: "user.php",
        type: "POST",
        data: formData,
        processData: false,
        contentType: false

});

上面的PHP代码可以很好地使用文件名直接调用操作，我得到了多个值，但是如果我使用ajax，那么它就不起作用了

下面是HTML代码

<form name="imgupload" enctype="multipart/form-data" method="post">
     <td rowspan="1">
        <input type="file" id="image1" name="image1[]" multiple>
    </td>
     <input type='submit' name='submit'><br/>
</form>

您需要表单提交事件才能让ajax按如下方式工作：

<form id="imgupload" name="imgupload" enctype="multipart/form-data" method="post">
    <td rowspan="1">
        <input type="file" id="txtImage1" name="images1[]" multiple>
    </td>
    <input type='submit' name='submit'><br/>
</form>
<script
    src="http://code.jquery.com/jquery-3.3.1.min.js"
    integrity="sha256-FgpCb/KJQlLNfOu91ta32o/NMZxltwRo8QtmkMRdAu8="
crossorigin="anonymous"></script>
<script>

    $('#imgupload').on('submit', function () {

        var formData = new FormData();            
        var count = document.getElementById('txtImage1').files.length;
        for (var x = 0; x < count; x++) {
            formData.append("image1[]", document.getElementById('txtImage1').files[x]);
        }

        $.ajax({
            url: "user.php",
            type: "POST",
            data: formData,
            processData: false,
            contentType: false

        });
        return false;
    });


</script>

$（'imgupload'）。在（'submit'，函数（）{
var formData=new formData（）；
var count=document.getElementById（'txtImage1'）.files.length；
对于（变量x=0；x

user.php

for($x=0; $x<count($_FILES['image1']['name']); $x++ ) {

    // echo "Hello";
    $file_name = $_FILES['image1']['name'][$x];

    $file_size = $_FILES['image1']['size'][$x];
    $file_tmp  = $_FILES['image1']['tmp_name'][$x];
    $lclLocation1 = "http://$_SERVER[HTTP_HOST]/images/".$_FILES["image1"]["name"][0];
    $lclLocation2 = "http://$_SERVER[HTTP_HOST]/images/".$_FILES["image1"]["name"][1];


    if(($_FILES["image1"]["type"][$x] == "image/gif") || ($_FILES["image1"]["type"][$x] == "image/jpeg") || ($_FILES["image1"]["type"][$x] == "image/png") || ($_FILES["image1"]["type"][$x] == "image/pjpeg")){
        $sumber = $file_tmp;
        $tujuan = "../images/" . $file_name;
        // echo "Image got";
        move_uploaded_file($sumber, $tujuan);

    }
    else  {
        echo "Only Images can be stored!";
    }
} // end for

---

for（$x=0；$x哪里是ajax调用？？？稍等，我将使用var关键字共享您声明的formData？应该是=>var formData=new formData（）；否，如下所示formData=new formData（）；请在使用它之前声明变量，这不是问题。文件只能通过文件输入访问，并且它是“上载”而不是“TXTMAGE1”，id为“TXTMAGE1”元素的元素在哪里？现在您将它更改为“image1”，并且您有了“TXTMAGE1”在javascript中，你在做什么？键入错误bro。我有两个程序是为调试而编写的，我在这里发布了旧程序。
for($x=0; $x<count($_FILES['image1']['name']); $x++ ) {

    // echo "Hello";
    $file_name = $_FILES['image1']['name'][$x];

    $file_size = $_FILES['image1']['size'][$x];
    $file_tmp  = $_FILES['image1']['tmp_name'][$x];
    $lclLocation1 = "http://$_SERVER[HTTP_HOST]/images/".$_FILES["image1"]["name"][0];
    $lclLocation2 = "http://$_SERVER[HTTP_HOST]/images/".$_FILES["image1"]["name"][1];


    if(($_FILES["image1"]["type"][$x] == "image/gif") || ($_FILES["image1"]["type"][$x] == "image/jpeg") || ($_FILES["image1"]["type"][$x] == "image/png") || ($_FILES["image1"]["type"][$x] == "image/pjpeg")){
        $sumber = $file_tmp;
        $tujuan = "../images/" . $file_name;
        // echo "Image got";
        move_uploaded_file($sumber, $tujuan);

    }
    else  {
        echo "Only Images can be stored!";
    }
} // end for