我可以使用 form_upload.php 将项目添加到 mysql 服务器并在服务器文件夹上上传照片，但照片无法显示在 view_data.php 上。我已经将文件权限设置为 777 但仍然无法工作。
这是我的 form_upload php 和 view_data.php:
<!DOCTYPE html>
<html>
<link href="style.css" rel="stylesheet" type="text/css">
<style type="text/css">
</style>
<title>Image Upload</title>
<body>
<div id="container">
<div class="content">
<form action="form_upload.php" method="post" enctype="multipart/form-data">
<fieldset>
<table width="350" border="0" align="center">
<legend>Image Information Entry
<tr><td><label>Title<span class="required">*</span></label></td>
<td align="center"><input type="text" name="title" placeholder="title"></br></td></tr>
<tr><td><label>Description<span class="required">*</span></label></td>
<td align="center"><input type="text" name="description" placeholder="description"></br></td></tr>
<tr><td><label>Username<span class="required">*</span></label></td>
<td align="center"><input type="text" name="username" placeholder="username"></br></td></td>
<tr><td><label>Mobile Number<span class="required">*</span></label></td>
<td align="center"><input type="text" name="mobilenumber" class="input-small" placeholder="mobilenumber"></br></td></tr>
<tr><td><label>Address<span class="required">*</span></label></td>
<td align="center"><input type="text" name="address" class="input-xlarge" placeholder="address"></br></td></tr>
<tr><td><label for="file">Upload Image:</label></td>
<td align="right"><input type="file" name="file" id="file"><br></td></tr>
<tr><td> </td></tr>
<tr><td colspan="2" align="center"><button type="submit" name="submit" class="btn">Submit</button>
<a href="view_data.php?o=0" class="btn btn-primary">View Gallery</a></td></tr>
<tr><td colspan="2">   </td></tr>
<tr><td colspan="2" align="center">
<?php
if(isset($_REQUEST['submit']))
{
$con=mysqli_connect("local","user","pass","table");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$allowedExts = array("gif", "jpeg", "jpg", "png");
$temp = explode(".", $_FILES["file"]["name"]);
$extension = end($temp);
$title=$_POST['title'];
$description=$_POST['description'];
$username=$_POST['username'];
$mobilenumber=$_POST['mobilenumber'];
$address=$_POST['address'];
$file=$_FILES["file"]["name"];
$size= $_FILES["file"]["size"];
if( empty($title) || empty($description) || empty($mobilenumber) || empty($address) || empty($file))
{
echo "<label class='err'>All field is required</label>";
}
elseif(!is_numeric($mobilenumber))
{
echo "<label class='err'>Mobile number must be numeric</label>";
}
elseif($size >40000)
{
echo "<label class='err'> Image size must not greater than 40kb </label>";
}
if ((($_FILES["file"]["type"] == "image/gif")
|| ($_FILES["file"]["type"] == "image/jpeg")
|| ($_FILES["file"]["type"] == "image/jpg")
|| ($_FILES["file"]["type"] == "image/pjpeg")
|| ($_FILES["file"]["type"] == "image/x-png")
|| ($_FILES["file"]["type"] == "image/png"))
&& ($_FILES["file"]["size"] < 40000)
&& in_array($extension, $allowedExts))
{
if ($_FILES["file"]["error"] > 0)
{
echo "Return Code: " . $_FILES["file"]["error"] . "<br>";
}
if (file_exists("/test/upload" . $_FILES["file"]["name"]))
{
echo $_FILES["file"]["name"] . "Image upload already exist. ";
}
else
{
move_uploaded_file($_FILES["file"]["tmp_name"],
"/home/tz005/public_html/test/upload/" . $_FILES["file"]["name"]);
mysqli_query($con,"INSERT INTO item_image (title, description, username, mobilenumber, address, filename)
VALUES ('$title', '$description', '$username', '$mobilenumber', '$address', '$file')");
echo "Image Information Successfully Saved!";
}
}
mysqli_close($con);
}
?>
</td></tr>
</legend>
</table>
</fieldset>
</form>
</div>
<div class="footer"></div>
</div>
</body>
</html>
<!DOCTYPE html>
<html>
<link href="style.css" rel="stylesheet" type="text/css">
<style type="text/css">
</style>
<title>view image information</title>
<body>
<div id="container">
<div class="con2">
<?php
$con=mysqli_connect("server","user","pass","table");
if (is_numeric($_GET['o']))
{
$o=$_GET['o'];
}else {
$o=0;
}
if ($o >=1){
$prev=$o-1;
} else{
$prev=0;
}
$query=mysqli_query($con,"SELECT * FROM item_image LIMIT $o, 1");
$get_pic=mysqli_fetch_assoc($query);
$query2=mysqli_query($con,"SELECT imageid FROM item_image");
$get_pic2=mysqli_fetch_assoc($query2);
$total=mysqli_num_rows($query2);
if ($o <=$total){$next=$o+1;}
?>
<?php do { ?>
<table align="center" width="300" border=".5" bordercolor="#0B615E">
<tr> <td colspan="2" align="center"><?php echo '<img src="/home/tz005/public_html/test/upload/filename' . $get_pic['filename'] . '" width="200" height="200"> '; ?></td></tr>
<tr><td width="60"> Details: </td> <td align="left" bordercolor="#0B615E"> <?php echo $get_pic['title']; ?>  
<?php echo $get_pic['description'];?>  
<?php echo $get_pic['username'];?> </td></tr>
<tr><td width="60"> Mobile number:</td> <td align="left"><?php echo $get_pic['mobilenumber']; ?></td></tr>
<tr><td width="60"> Address: </td> <td align="left"><?php echo $get_pic['address']; ?></td></tr>
<tr><td colspan="2" align="center">
<?php
} while ($get_pic=mysqli_fetch_assoc($query));
?>
<?php if ($o>0){ ?>
<span><a href="view_data.php?o=<?php echo $prev; ?>">Previous</a></span>
<?php } ?>
<?php if ($o < ($total - 1)){ ?>
<span><a href="view_data.php?o=<?php echo $next; ?>">Next</a></span>
<?php } ?>
<?php
mysqli_close($con);
?>
</td><tr>
<tr><td colspan="2">   </td></tr>
<tr>
<td colspan="2" align="center"><a href="form_upload.php"> Back to Image Information Entry</td></tr>
</table>
</div>
</div>
</body>
</html>
最佳答案
您的图像标签指向您的文件系统，它需要指向一个 URL:
<img src="/home/tz005/public_html/test/upload/filename' . $get_pic['filename'] . '"
您可能应该将其更改为:
<img src="/test/upload/'.$get_pic['filename'].'"
关于php - 网页上的照片破裂，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/23751079/