Question

请一些人可以帮助我，我是新的php我试图显示存储在一个名为uploads的文件夹中的图像，所有的图像路径都存储在我调用的一个字段中，文件 这是表单代码

<form enctype="multipart/form-data" action="" method="post">
    <table width="800" align="center">
<tr>
<td> <h1>Insert new post</h1> </td>
</tr>
<tr>
<td>Post Title:</td>
<td><input type="text" name="post_title" /></td>
</tr>
<tr>
<td>Post Image:</td>
<td> 
<input name="file[]" type="file" id="file"/>
<input type="button" id="add_more" class="upload" value="Add More Files"/> 
</td>
    </tr>

    <tr>
    <td>Post Content:</td>
<td><textarea name="post_content" rows="16" cols="30"> </textarea></td>
    </tr>
<tr>

 <?php 

 if (isset($_POST['submit'])) {

 $errors= array();

 $j = 0;     
$target_path = "uploads/";     

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

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

 $ext = explode('.', basename($_FILES['file']['name'][$i]));  

$file_extension = end($ext); // Store extensions in the variable.

 $target_path = $target_path. md5(uniqid()) . "." . $ext[count($ext) - 1] ;   

 $j = $j + 1;    

 if (($_FILES["file"]["size"][$i] < 100000)    
 && in_array($file_extension, $validextensions)) {
  if (move_uploaded_file($_FILES['file']['tmp_name'][$i], $target_path)) {
 $file_name_all.=$target_path."*";
            $filepath = rtrim($file_name_all.''); 

       $post_title = $_POST['post_title'];
 $post_content = $_POST['post_content'];

if($post_title=='') {

 echo "<script>alert('please fill all feilds') </script>" ;
        exit();
        }   

    else {   

     $insert_posts = "insert into posts              (post_title,file,post_content) values          ( '$post_title','$target_path','$post_content')";

      $run_posts = mysql_query($insert_posts);

    echo"<script>alert('post has been published!')</script>";
    } 
} 
    ?>

，这是预览代码

<?php
    if(!isset($_GET['cat'])) {

    $get_posts = "select *from posts limit 0,5 ";
    $run_posts = mysql_query($get_posts);
    while ($row_posts = mysql_fetch_array($run_posts))

    $post_id = $row_posts['post_id'];               
    $post_title = $row_posts['post_title'];
    $file =   $row_posts['file'];
    $post_content =substr($row_posts['post_content'],0,200);

    echo"
    <li>
<h3><a id=1title' href='details.php?post=$post_id'>$post_title</a></h3>
<img  src='uploads/$file' width='250' height='150'/>    
</p>    
$post_content 
<div >

 </div><br/> 

</li>       
    ";
 ?>

以下是我上传到数据库的两张图片的路径，我的目的是显示这些图片

上传/ c1d225856af139c794b5358e61e99cd3.jpg1935ecdda8ed0c5602286a5398959931.jpg

如何从数据库路径回显数组中的所有图像

0 个答案: