i am getting error in php while uploading image to databse using jquery, the data in variables arent passing maybe

前端 未结 3 2023
余生分开走
余生分开走 2021-01-20 12:29

My code is a form where it picks a file from the user, then send the data using jQuery to a PHP file where it gets the image content and displays it and in a success functio

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3条回答
  • 2021-01-20 12:41

    First of all insert your input file tag in a form and use enctype="multipart/formdata" to send an image otherwise you will not able to send image

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  • 2021-01-20 12:51

    This should work!

    HTML:

    <form id="my-upload-form" method="post" enctype="multipart/form-data">
    <input type="file" name="required-image" />
    <button> Upload </button>
    </form>
    

    JS:

    $("button").click(function(e) {
    /* prevent default form action */
    e.preventDefault();
    /* get form element */
    var formElement = document.getElementById("my-upload-form");
    /* collect all form data from Form element */
    var formData = new FormData(formElement);
    
    $.ajax({
    url: '/path-to-form-handler.php',
    data: formData,
    cache: false,
    contentType: false,
    processData: false,
    method: 'POST',
    success: function(response) {
        console.log(response);
    }
    });
    });
    

    PHP:

    <?php
    
    /* for this example, $_FILES["required-image"] would be an array having image details */
    echo $_FILES["required-image"]["name"];
    echo $_FILES["required-image"]["type"];
    echo $_FILES["required-image"]["tmp_name"];
    echo $_FILES["required-image"]["size"];
    
    ?>
    
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  • 2021-01-20 13:06

    Approach

    You need to use new FormData() object.

    The FormData interface provides a way to easily construct a set of key/value pairs representing form fields and their values, which can then be easily sent using the XMLHttpRequest.send() method. It uses the same format a form would use if the encoding type were set to "multipart/form-data".

    So you don't actually have to declare a form tag and add inputs inside, yes it makes it easier if you have let us make a call assuming that you do not have a form tag.

    Problem

    The problem in your script is that your formdata is a json rather than a FormData() interface object, which uses formdataObject.append() which appends a new value onto an existing key inside a FormData object, or adds the key if it does not already exist.

    See code below which posts email, file label and a file to a PHP page without using form tag for the inputs.

    Without <form> tag

    Assuming that your html looks like below without a form

    <label>Your email address:</label>
    <input type="email" autocomplete="on" autofocus name="userid" placeholder="email" required size="32" maxlength="64" />
    <br />
    <label>Custom file label:</label>
    <input type="text" name="filelabel" size="12" maxlength="32" />
    <br />
    <label>File to stash:</label>
    <input type="file" name="file" required />
    <input type="button" name="submit" value="Stash the file!" />
    

    Your javascript code will look like below

    $(document).ready(function () {
        $("input[name='submit']").on('click', function (event) {
    
            event.preventDefault();
    
            //START Append form data
            var data = new FormData();
            data.append(
                'userid', $("input[name='userid']").val());
            data.append(
                'label', $("input[name='filelabel']").val()
            );
            data.append('file', $("input[name='file']")[0].files[0], 'somename.jpg');
            //END append form data
    
            $.ajax({
                type: "POST",
                url: "file.php",
                data: data,
                processData: false,
                contentType: false,
    
                success: function (data) {
                    console.log("SUCCESS : ", data);
                },
                error: function (e) {
                    console.log("ERROR : ", e);
                }
            });
    
        });
    
    });
    

    And your file.php will look like below

    <?php
    
    print_r($_POST);
    print_r($_FILES);
    

    This should show you the file inputs and file both of them in the console when you hit the stash file button.

    With <form> tag

    If you have the inputs wrapped inside the form tag then your code will be changed on the following sections

    • Change binding of click event to form submit event.

    • Change button type to submit in the HTML.

    • Get the form object.

    • Use form object to initialize the FormData().

    See below How your HTML will look like

    <form enctype="multipart/form-data" method="post" name="fileinfo">
        <label>Your email address:</label>
        <input type="email" autocomplete="on" autofocus name="userid" placeholder="email" required size="32" maxlength="64" />
        <br />
        <label>Custom file label:</label>
        <input type="text" name="filelabel" size="12" maxlength="32" />
        <br />
        <label>File to stash:</label>
        <input type="file" name="file" required />
        <input type="submit" value="Stash the file!" />
    </form>
    

    And your javascript will look like below

    $(document).ready(function () {
        $("form").on('submit', function (event) {
    
            event.preventDefault();
            var form = this;
            var data = new FormData(form);
    
            $.ajax({
                type: "POST",
                url: "file.php",
                data: data,
                processData: false,
                contentType: false,
    
                success: function (data) {
                    console.log("SUCCESS : ", data);
                },
                error: function (e) {
                    console.log("ERROR : ", e);
                }
            });
    
        });
    
    });
    
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