Question

我有一个表单，有助于将多个图像上传到服务器文件夹并将其路径保存在数据库中，问题是所有图像的路径都上传到一列，由于它我遇到了问题获取显示图像的路径。解决方案是我要么让图像存储在不同的列中，要么找到分离图像的方法，但是我无法做到这两种。

存储的路径类似于

uploads/c376437e2a45598b2f4d89eae4f191e8.png*uploads/c376437e2a45598b2f4d89eae4f191e8.png8069756be5095978123ae51fadbffe3b.png*uploads/c376437e2a45598b2f4d89eae4f191e8.png8069756be5095978123ae51fadbffe3b.png04aaa414c21dc057bc594b896124068e.png

数据库结构演示

id  offimage offimage1   offimage2

到目前为止我的代码（表单和php代码在不同的页面上）

<form enctype="multipart/form-data" action="co_insert_office_image.php?id=<?php echo $_GET['id']; ?>" method="post">
        First Field is Compulsory. Only JPEG,PNG,JPG Type Image Uploaded. Image Size Should Be Less Than 100KB.
    <hr/>
    <div id="filediv"><input name="file[]" type="file" id="file"/></div><br/>
    <input type="button" id="add_more" class="upload" value="Add More Files"/>
    <input type="submit" value="Upload File" name="submit" id="upload" class="upload"/>
</form>

co_insert_office_image.php

<?php
ob_start();
include('co_session.php');
$con=mysqli_connect("localhost","root","","db");
// Check connection
if (mysqli_connect_errno()) {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
if (isset($_POST['submit'])) {
    $j = 0; //Variable for indexing uploaded image 

    $target_path = "uploads/"; //Declaring Path for uploaded images
    for ($i = 0; $i < count($_FILES['file']['name']); $i++) {//loop to get individual element from the array

        $validextensions = array("jpeg", "jpg", "png");  //Extensions which are allowed
        $ext = explode('.', basename($_FILES['file']['name'][$i]));//explode file name from dot(.) 
        $file_extension = end($ext); //store extensions in the variable

        $target_path = $target_path . md5(uniqid()) . "." . $ext[count($ext) - 1];//set the target path with a new name of image
        $j = $j + 1;//increment the number of uploaded images according to the files in array       

      if (($_FILES["file"]["size"][$i] < 100000) //Approx. 100kb files can be uploaded.
                && in_array($file_extension, $validextensions)) {
            if (move_uploaded_file($_FILES['file']['tmp_name'][$i], $target_path)) {//if file moved to uploads folder
                //echo $j. ').<span id="noerror">Image uploaded successfully!.</span><br/><br/>';

                $file_name_all.=$target_path."*";
                $filepath = rtrim($file_name_all, '*'); 
                //echo $filepath;
                $officeid = $_GET['id'];
                $sql = "UPDATE register_office SET offimage='$filepath' WHERE id='$officeid' ";
                            if (!mysqli_query($con,$sql)) 
                                {
                                    die('Error: ' . mysqli_error($con));
                                }

            } else {//if file was not moved.
                echo $j. ').<span id="error">please try again!.</span><br/><br/>';
            }
        } else {//if file size and file type was incorrect.
            echo $j. ').<span id="error">***Invalid file Size or Type***</span><br/><br/>';
        }
    }
    header("Location: co_request_sent.php ");
}
mysqli_close($con); 
?>

Answer 1

最后我得到了解决方案，我看到有些人标记了这个问题，他们可能正在寻找答案，所以这里是

首先连接数据库，获取数据然后使用以下代码

while($row1 = mysqli_fetch_assoc($result1)) 
    {
    $str= $row1["offimage"];
    //print_r (explode("*",$str));

    $array =  explode('*', $str);
    foreach ($array as $item) {
        //echo "<li>$item</li>";
        echo "<img src=\"http://localhost/cms/1/". $item . "\" height=\"200\" width=\"200\"/>";
    }

}

将图像路径存储在不同的数据库列中

1 个答案: