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php mysql 从数据库中检索上传的照片

转载 作者：行者123 更新时间：2023-11-29 08:32:23 25 4
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我试图将用户列表的搜索结果显示为与从数据库检索到的相同结果，但系统显示路径而不是图像本身。谁能帮我吗？

到目前为止，我都是从按名称搜索的查询开始的:

搜索.php

<?php
session_start();
if($_SESSION['login'] != 'true'){
        header("location:index.php");
    }


   $login = ($_SESSION['login']);
   $userid = ($_SESSION['user_id']);
   $login_user = ($_SESSION['username']);
   $fname = ($_SESSION['first_name']);
   $lname = ($_SESSION['last_name']);
   $sessionaddres =($_SESSION['address']);


require_once('for members/scripts/connect.php'); 

// function for selecting names
function nameQuery(){

  $nameData = mysql_query("SELECT * FROM user") or die("could select database");

  while($record = mysql_fetch_array($nameData)){

     echo'<option value="' . $record['user_name'] .  '">' . $record['user_name'] . '</option>';

  }

}



// function for select by specialization

function specializationQuery(){

$specData = mysql_query("SELECT * FROM specialization");

  while($recordJob = mysql_fetch_array($specData)){

     echo'<option value="' . $recordJob['specialization_name'] .  '">' . $recordJob['specialization_name'] . '</option>';

  }


}

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


    $Sgov = $_POST['governorate'];
    @$Sdist = $_POST['district'];
    @$Svillage = $_POST['village'];


    // query  search by name
    if(isset($_POST['name']))
    {
        $Sname =$_POST['name'];
    $sql = mysql_query("SELECT first_name, last_name, profile_pic FROM user WHERE user_name ='$Sname'")or die(mysql_error());

    if($sql)
    {
     while($getrow = mysql_fetch_array($sql))
     {
        $firstname = $getrow['first_name'];
        $lastname = $getrow['last_name'];
        $profilepic = $getrow['profile_pic'];

        if($profilepic == "") 
         {
             $profile_pic = "images/default_img.jpg";
         }
         else
         {
             $profile_pic = "userdata/profile_pics/".$profilepic;
         }



         echo "<ul>
          <li>
           ".$firstname. " ".$lastname ."
          </li>
          <li>
           <img style='width:80px' src='".$profile_pic . "'>
          </li>
         </ul>";
     }
   }
   else
   {
       echo "their was no result!!!!";
   }
  }
}

// search by specialization

if(isset($_POST['specialization']))
{
    $Sspec = $_POST['specialization'];
    $sql = mysql_query("first_name, last_name, profile_pic FROM user WHERE specialization = '$Sspec'")or die(mysql_error());
    if($sql)
    {
     while($getrow = mysql_fetch_array($sql))
     {
        $firstname = $getrow['first_name'];
        $lastname = $getrow['last_name'];
        $profilepic = $getrow['profile_pic'];

        if($profilepic == "") 
         {
             $profile_pic = "images/default_img.jpg";
         }
         else
         {
             $profile_pic = "userdata/profile_pics/".$profilepic;
         }



         echo "<ul>
          <li>
           ".$firstname. " ".$lastname ."
          </li>
          <li>
           <img style='width:80px' src='".$profile_pic . "'>
          </li>
         </ul>";
     }
   }
   else
   {
       echo "their was no result!!!!";
   }

}
?>


<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>search page</title>
<link href="style/stylesheet.css" rel="stylesheet" type="text/css" />
<script type = "text/javascript" src = "http://ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script>

<script type="text/javascript">

 $(document).ready(function(){
                $("select#district").attr("disabled","disabled");
                $("select#village").attr("disabled","disabled");
                $("select#governorate").change(function(){
                $("select#district").attr("disabled","disabled");
                $("select#district").html("<option>wait...</option>");
                var id = $("select#governorate option:selected").attr('value');
                $.post("select_district.php", {id:id}, function(data){
                    $("select#district").removeAttr("disabled");
                    $("select#district").html(data);
                });
            });
            $("select#district").change(function(){ 
            id = $(this).val(); 
            $("select#village").attr("disabled","disabled");
            $("select#village").html("<option>wait...</option>");
            $.post("select_village.php", {id:id}, function(data){


            $("select#village").removeAttr("disabled");
            $("select#village").html(data);
         });
        });
            $("form#registerform").submit(function(){
                var cat = $("select#governorate option:selected").attr('value');
                var type = $("select#district option:selected").attr('value');
                var village = $("select#village option:selected").attr('value');

            });
    });


</script>
</head>


<body>
<div class="container">
<!--<?php require_once('header.php'); ?>-->
<br />
<br />
<br />
<br />

  <!-- <?php require_once('leftsideBar2.php'); ?>-->


   <div id="search-title">Search section</div>
    <div id="search-form">
    <?php include "select.class.php"; ?>
        <form action="search.php" method="post">

         Search By Name:<br />
         <select name="name" >
            <?php nameQuery(); ?>
             <option id="0">-- select By UserName --</option>
             </select>
           <br/><br/>
           Search By Governorate:<br />

        <select id="governorate" name = 'governorate'>
                <?php echo $opt->ShowGovernorate(); ?>
                </select>
           <br /><br/>
           Search by District:<br />

      <select id="district" name="district">
                    <option value="0">choose...</option>
                </select>

           <br /><br/>
           Search by Cities:<br />
           <select id="village" name="village">
                    <option value="0">choose...</option>
                </select>
           <br /><br/>
           Search By Specialization:<br />
           <select name="specialization">
             <option id="0" disabled="disabled">-- select Job --</option>

          <?php specializationQuery(); ?>
          </select>
          <input type="submit" name="search" value="Search"  />
        </form>

    </div>
</div>
<?php require_once('footer.php'); ?>
</body>
</html>

最佳答案 

   echo "<ul>
          <li>
           ".$firstname. " ".$lastname ."
          </li>
          <li>
           <img src='".$profile_pic . "'>
          </li>
         </ul>";

关于php mysql 从数据库中检索上传的照片，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/16243771/

25 4 0
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