Question

我在开发的网站上显示图像时遇到问题。这是一个网站，用户可以更改他们的个人资料图片，以及他们的基本个人资料信息。下面是我的示例代码。

profile.php

      <?php 
                    $studpix=$row_rsp['pix'];
                    $propix='<img class=
   "profile-user-img img-responsive img-circle" src="...
    /Student /imageupload/blank.png"
                             alt="profile picture">';

                      if($propix!=NULL)
                      {
                    $propix='<img class="profile-user-img 
    img-responsive img-circle" src="Student/imageupload/'.$studpix.'"
                             alt="profile picture">';

                        };
                        $profile_pic_btn = 
'<a href="#"   onclick="clicked(\'avatar_form\')">  Profile pics</a>';
$avatar_form  = '<form id="avatar_form" 
enctype="multipart/form-data"   method="post" action="photoup.php">';
$avatar_form .=   '<h4>Change your picture</h4>';
$avatar_form .=   '<input type="file" name="avatar" required>';
$avatar_form .=   '<p><input type="submit" value="Upload"></p>';
$avatar_form .= '</form>';
                        ?>
  <?php echo $propix?><?
   php echo $avatar_form?><?php   echo $profile_pic_btn;?> 


  //other codes goes here

imageupload.php

               <?php 
if (isset($_FILES["avatar"]["name"]) && $_FILES["avatar"]
["tmp_name"] != ""){
$fileName = $_FILES["avatar"]["name"];
$fileTmpLoc = $_FILES["avatar"]["tmp_name"];
$fileType = $_FILES["avatar"]["type"];
$fileSize = $_FILES["avatar"]["size"];
$fileErrorMsg = $_FILES["avatar"]["error"];
$kaboom = explode(".", $fileName);
$fileExt = end($kaboom);
list($width, $height) = getimagesize($fileTmpLoc);
if($width < 10 || $height < 10){
    echo "ERROR: That image has no dimensions";
    exit(); 
}
$db_file_name = rand(100000000000,999999999999).".".$fileExt;
if($fileSize > 1048576) {
    echo "ERROR: Your image file was larger than 1mb";
    exit(); 
} else if (!preg_match("/\.(gif|jpg|png)$/i", $fileName) ) {
    echo "ERROR: Your image file was not jpg, gif or png type";
    exit();
} else if ($fileErrorMsg == 1) {
    echo "ERROR: An unknown error occurred";
    exit();
}
$sql = "SELECT pix FROM studentdetails WHERE email='%s'";
$query = mysqli_query($myconn, $sql);
$row = mysqli_fetch_row($query);
$avatar = $row[0];
if($avatar != ""){
    $picurl = "../Student/imageupload/$avatar"; 
    if (file_exists($picurl)) { unlink($picurl); }
}
$moveResult = move_uploaded_file(
 $fileTmpLoc, "../Student  /imageupload       /$db_file_name");
if ($moveResult != true) {
    echo "ERROR: File upload failed";
    exit();
}
include_once("../image_resize.php");
$target_file = "../Student/imageupload/$db_file_name";
$resized_file = "../Student/imageupload/$db_file_name";
$wmax = 200;
$hmax = 300;
img_resize($target_file, $resized_file, $wmax, $hmax, $fileExt);
$sql = "UPDATE studendetails SET pix='%s' WHERE email='%s' LIMIT 1";
$query = mysqli_query($myconn, $sql);
mysqli_close($myconn);
header("location: profile.php");
exit();
}
?>

我们将不胜感激。

Answer 1

尝试逐步调试代码。

1。检查文件是否正确上传到正确的文件夹。

2. 检查数据库中的数据是否正确更新。

3. 尝试直接在浏览器中通过网址打开文件。

4. 检查您的HTML代码是否在网页上正确输出并调试输出的源代码。

5. 确保您的HTML代码正常运行。

可能还有更多步骤，但这可能会给你一些指导。

图像不在PHP中以我的形式显示

1 个答案: