Question

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

这是我的ajax查询

  // this contains two images
var lclImage1 = document.getElementById("txtImage1");

    lclImage2 = lclImage1.files[0];

    var formData = new FormData();

    formData.append("image1", lclImage2);

此后，我们将ajax发送到PHP。

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

});

上面的PHP代码可以很好地用于直接使用文件名调用操作，我得到多个值，但是如果我使用ajax，则它不起作用

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>

Answer 1

您将需要表单提交事件以使ajax如下工作，

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

user.php

tidyverse

我还编辑了图像扩展条件。由于文件名的名称中可能带有句点，因此使用句点将文件名爆炸会导致错误。

无法从PHP中的ajax访问2个或更多文件

1 个答案: