2011-02-24 267 views
1

我在项目中的图片中有一个默认图像,但似乎无法让它反映出用户何时未上传图像。请帮助。下面是纵断面图,并上传脚本如何设置默认的默认图像(no_pic.gif)php?

Profile.php:

<head> 
     <title>Profile</title> 
    </head> 
    <body> 
     <table> 
      <tr> 
       <td colspan="2">View Your Current Profile</td> 
      </tr> 
      <tr> 
       <td>Title:</td> 
       <td><?php echo $jobseekertitle; ?></td> 
      </tr> 
      <tr> 
       <td>First Name:</td> 
       <td><?php echo $jobseekerfirstname; ?></td> 

      </tr> 
      <tr> 
       <td>Last Name:</td> 
       <td><?php echo $jobseekerlastname; ?></td> 
      </tr> 
      <tr> 
       <td>Address:</td> 
       <td><?php echo $jobseekeraddress; ?></td> 
      </tr> 
      <tr> 
       <td>Phone:</td> 
       <td><?php echo $jobseekerphoneno; ?></td> 
      </tr> 
      <tr> 
       <td>D.O.B (mm/dd/yyyy):</td> 
       <td><?php echo $jobseekerdob; ?></td> 
      </tr> 
      <tr> 
       <td>Gender:</td> 
       <td><?php echo $jobseekergender; ?></td> 
      </tr> 
      <tr> 
       <td>Picture:</td> 

       <td><?php echo "<img src='$location' width='100' height='100'>";?></td> 
      </tr> 
     </table> 

     <br><a href='upload.php'> Image upload</a> 

     <br><a href='resumecreator.php'> Resume information</a> 

     <br><a href='jobpreference.php'> Job Preference</a> 

</body> 
</html> 

upload.php的

<?php 
// form in which we can upload the image 
session_start(); 

include("connect.php"); 

// session as on website. 

$jobseekerid = $_SESSION['jobseekerid']; 

if($_POST['submit']) 
{ 
    //get file attributes 
    $name = $_FILES['myfile']['name']; 
    $tmp_name = $_FILES['myfile']['tmp_name']; 


    if ($name) 
    { 
     // start upload process 
     $location ="pictures/$name"; 
     move_uploaded_file($tmp_name,$location); 
// updating table users (setting image locaion 
     $query = mysql_query("UPDATE jobseekers SET imagelocation='$location' WHERE jobseekerid='$jobseekerid'"); 

     die("Your avatar has been uploaded! <a href='profile.php'>Go back to view</a>"); 


    } 
    else 

     die("Please Select a file!"); 

} 

echo "Welcome, ".$_SESSION['username']."!<br><a href='logout.php'>Logout</a>"; 

echo "Upload Your image: 

<form action='upload.php' method='POST' enctype='multipart/form-data'> 
File: <input type='file' name='myfile'> <input type='submit' name='submit' value='upload!' 
> 
</form> 

"; 

?> 
+0

当我需要显示图像的东西,我检查数据库图像列,如果长度小于一定的数字我的服务器代码应用占位符图像,否则图像在数据库中。 – 2011-02-24 23:10:35

回答

4

我通常不喜欢这样

<td><img src="<?= @$location ? $location : 'default/image.gif' ?>" width="100" height="100" /></td> 

这样如果$location不存在,它将抑制错误并输出默认替换。

+0

你好,刚刚尝试过,但它仍然只是显示一个破碎的图像。默认图片是no_pic.gif。有关如何解决这个问题的更多想法。​​ ope 2011-02-24 23:25:02

+0

并且结果是? – jondavidjohn 2011-02-24 23:25:59

+0

短标签**和**错误抑制?ick。 – drudge 2011-02-24 23:26:13