Question

这是显示图像和消息的代码。它以二进制显示图像，不显示消息。此外，我无法将从我的Android应用程序发送的图像存储到数据库。

    <?php
    error_reporting(E_ALL);
    ini_set("display errors",1);
    $servername = "localhost";
    $username = "root";
    $password = "raspberrypi3";
    $dbname = "board2";
    $conn = new mysqli($servername, $username, $password, $dbname);
    if ($conn->connect_error)
    {
    die("Connection failed: " . $conn->connect_error);
    } 
    if($_SERVER['REQUEST_METHOD'] == 'POST')
    {
    $DefaultId =1;
    $div=$_POST["div"];
    $message=$_POST["data"];
    $image = $_POST['image'];
    // $GetOldIdSQL ="SELECT id FROM userinfo ORDER BY id ASC";
    $ImagePath ="C:\\wamp\\www\\project3\\image\\$DefaultId.png";
    $ServerURL = "http://www.localhost.com//$ImagePath";
    //$ServerURL ="$DefaultId.png";
    if($div=="2"){
    $InsertSQL = "UPDATE userinfo SET `image` = '$ServerURL' WHERE `div` = $div";          
    }else{
    $InsertSQL = "UPDATE userinfo SET `message` = '$message' WHERE `div` = $div";          
    }
    // $InsertSQL = "INSERT INTO userinfo (`message`,`div`,`image`) VALUES('$message','$div','$ServerURL')";
    if(mysqli_query($conn, $InsertSQL)){
    if($div=="2"){
    file_put_contents($ImagePath,base64_decode($image));
    }
    echo "Process done.";
    } 
   mysqli_close($conn);
     }else{
   echo "Please Try Again";      }
   ?>

这是从Android应用程序接收代码并存储在数据库中的代码。

<?php
error_reporting(E_ALL);
ini_set("display errors",1);
?>
<html>
<head><h1><b><center>DIGITAL NOTICE BOARD</center></b></h1><br/><br/></br>

<script type="text/JavaScript">
        <!--
          function AutoRefresh( t ) {
           setTimeout("location.reload(true);", t);
        }
    //-->
   </script>

  <body onload="JavaScript:AutoRefresh(5000);" style="background-image:url(11.jpg) ;background-size: 1600px 1200px;box-shadow: inset 0 0 0 0px black;
    background-color: rgb(224, 141, 169);"height="1600" width="1200">
    <?php
    $servername = "localhost";   
    $username = "root";
    $password = "raspberrypi3";
    $dbname = "board2";
    $conn = new mysqli($servername, $username, $password, $dbname);
    if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
    } 
    $sql1 = "SELECT * from  userinfo where `div`=1" ;
$result1 = mysqli_query($conn, $sql1) or die(mysqli_error($conn)) ;
if ($rows1 = mysqli_fetch_array($result1))
{ 

?>
 <div  id="1";
 style="
 width: 66%;
  height: 379px;
 border: aqua:82;
  box-shadow: 0px 0px 5px #000;
    position : absolute ;
    background-image: url(download.png);
    ">
   <h1><b><font style="font-family: cursive;"><?php   echo $rows1['1']
  ?></h1></b></font>

  </div>
  <?php
  } 


  $sql2 = "SELECT * from  userinfo where `div`=2" ;
  $result2 = mysqli_query($conn, $sql2) or die(mysqli_error($conn)) ;
  if ($rows2 = mysqli_fetch_array($result2))
  {

   ?>
  <div style="
   width: 31%;
   height: 380px;
   border: aqua:82;
   box-shadow: 0px 0px 5px #000;
    position : absolute ;
    left : 68%;

   ">


   <img src="image/1.png "  height="380" width="425"/>


     </div>

 <?php
 } 

  $sql3 = "SELECT * from  userinfo where `div`=3" ;
  $result3 = mysqli_query($conn, $sql3)or die(mysqli_error($conn)) ;
  if ($rows3 = mysqli_fetch_array($result3))

  {
  ?>
  <div style="
   width: 97%;height: 73px;
   border: aqua:82;
   box-shadow: 0px 0px 5px #000;
    position : absolute ;
    left : 1% ;
    top : 85% ;
    background-image: url(maroon.png);
    ">

    <marquee><h1><b><font style="font-family: cursive;" font color="white"><?php   echo $rows3['1']
  ?></marquee></h1></b></font>
    </div>
    <?php
    }
     $conn->close();

    //echo "Hello,Android__";
   //$msg=urldecode($_POST['POST']);
   //echo $msg;
   ?>

Answer 1

如果您的图片是Blob（我不知道），您可以这样做：

<?php 
$connect = mysqli_connect("localhost", "user", "pass", "db");
$query = "SELECT * FROM table";

if(mysqli_num_rows($image) > 0) { 
while($row = mysqli_fetch_array($image)) { 

echo 
'<img src="data:image/jpeg;base64,'.base64_encode($row['image'] ).'" alt="image" class="any class">';
echo 
''$row['message'] ).'';

}} else { echo 'Error on connection';}?>

如果这是您所需要的，现在要存储您需要的图像：

$file = addslashes(file_get_contents($_FILES["image"]["tmp_name"]));
$query = "INSERT INTO db_table (image) VALUES ('".$file."')"; // query to insert

这应该可以解决你的问题

我试图从phpmyadmin数据库中显示图像和消息

1 个答案: