如果没有插入图像,则不更新字段

7vux5j2d  于 2021-06-20  发布在  Mysql
关注(0)|答案(2)|浏览(321)

我对php还很陌生,但是我在编写一些代码的过程中遇到了一个问题。
我正试图让下面的工作,以便用户可以上传两个图像在一个窗体,上传到服务器,并更新sql字段,但我有一个困难的时候,如何使它不更新sql字段,除非一个图像上传-我已经设法使它与一个图像使用;

$uploadArtwork = $_FILES['asset_name']['tmp_name'];

if($uploadArtwork == null) {
$sql = "";
}

else {
$sql = ""; 
}

我正在努力想,我如何才能做到这两个图像(最终超过两个图像?)
尝试了很多的谷歌搜索,但没有太多的运气呢!

vsikbqxv

vsikbqxv1#

$uploadArtwork1 = $_FILES['asset_name1']['tmp_name'];
$uploadArtwork2 = $_FILES['asset_name2']['tmp_name'];

// Image1 and/or image2 was uploaded successfully
if(($uploadArtwork1 != null) || ($uploadArtwork2 != null)) {
    $sql = "";
// No images were selected, or there were problems uploading them
} else {
    $sql = ""; 
}

尽管最好检查$\u files['asset\u name']['error']==upload\u err\u ok以确定是否成功上载了图像:

$uploadArtwork1 = $_FILES['asset_name1']['error'];
$uploadArtwork2 = $_FILES['asset_name2']['error'];

// Image1 and/or image2 was uploaded successfully
if(($uploadArtwork1 == UPLOAD_ERR_OK) || ($uploadArtwork2 == UPLOAD_ERR_OK)) {
    // Do something with $_FILES['asset_name1']['tmp_name'] and $_FILES['asset_name2']['tmp_name']
    $sql = "";
// No images were selected, or there were problems uploading them
} else {
    $sql = ""; 
}

更新:

require_once("Inc/classCloud.php");

$sql = "UPDATE assets SET asset_title='$post_asset_title'";

if ($uploadArtwork != null) {

    $getImageID= $res['data'];
    $sql .= ", asset_name='$getImageID'";

}

if ($uploadMock != null) {

    $getImageID2= $res2['data'];
    $sql .= ", product_artwork='$getImageID2'";

}

$sql .= " WHERE asset_id='$post_asset_id'";
yqhsw0fo

yqhsw0fo2#

这里有一个基本的结构。
基本上是循环浏览所有上传的文件,如果找到了,就把它们移到服务器上的一个新位置,并将条目写入数据库。
此代码尚未测试。

<?php

// Loops through all possible file uploads.
foreach ($_FILES as $file) {

    // Checks a file has been chosen.
    if (isset($file['tmp_name']) && !empty($file['tmp_name'])) {

        // Checks the uploaded (object) is a file.
        if (is_file($file['tmp_name'])) {

            // The filepath for the uploaded file.
            $destination = 'LOCATION TO MOVE THE UPLOADED FILE TO';

            /* 
             * Perform SQL Write here
             */

            if (WRITE WAS SUCCESSFUL) {

                // Move FIle
                move_uploaded_file($file['tmp_name'], $destination);    
            }
        }
    }
}

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