Question

声明

会话消息未显示在主页（main.php）

假定包含 connection.php 。（建立与MySQL服务器的连接）

我尝试过的事情

根据 add.php 中的以下代码，我有if-else语句来检查$url = mysqli_real_escape_string($conn,$_POST['url']);为UNIQUE的重复输入。但是在我尝试使用prepared statement之后，却没有显示会话消息。

main.php

<?php require_once 'add.php';
      include 'connection.php'
?>

<?php if (isset($_SESSION['message'])): ?>
     <div class="alert alert-<?=$_SESSION['msg_type']?> alert-dismissible fade show" role="alert">
          <button type="button" class="close" data-dismiss="alert"><span style='font-size:25px;'>&#215;</span></button>
     <?php

          echo $_SESSION['message'];
          unset($_SESSION['message']);
     ?>
     </div>
<?php endif ?>

add.php

if(isset($_POST['Save']))
      {
          $category = mysqli_real_escape_string($conn,$_POST['category']);
          $url = mysqli_real_escape_string($conn,$_POST['url']);
          $datetime = mysqli_real_escape_string($conn,$_POST['datetime']);
          $lat = mysqli_real_escape_string($conn,$_POST['lat']);
          $lng = mysqli_real_escape_string($conn,$_POST['lng']);

          $stmt = $conn->prepare("INSERT INTO crimenews (crimenews_cat, crimenews_url, crimenews_datetime, crimenews_locationLat, crimenews_locationLong) VALUES ('$category', '$url', '$datetime', '$lat', '$lng') ON DUPLICATE KEY UPDATE crimenews_url = VALUES(crimenews_url)");
          $stmt->bind_param("sssss",$_POST['category'],$_POST['url'],$_POST['datetime'],$_POST['lat'],$_POST['lng']);
          $stmt->execute();

          $result = $stmt->get_result();

          if($result->num_rows>0)
          {
              //Fetch URL on each row
               while($row = $result->fetch_assoc())
               {
                    $crimenews_url = $row["crimenews_url"];
               }
               //If there is no duplicated URL inserted into MySQL server
               if($crimenews_url == $url)
               {
                   $_SESSION['message'] = "The news has added.";
                   $_SESSION['msg_type'] = "success";
               }
               //If there is duplicated URL insert into MySQL server
               else
               {
                   $_SESSION['message'] = "The news has already existed.";
                   $_SESSION['msg_type'] = "danger";
               }
           }

           $stmt->close();

           header("location: main.php");
        }

预期结果

会话消息显示在main.php中。

Answer 1

您的sql INSERT语句完全不正确-这很可能是代码失败的原因。当您使用prepared statement并尝试将bind的值赋给占位符时，您实际上必须具有占位符才能起作用。

$sql='INSERT INTO `crimenews`
        (crimenews_cat, crimenews_url, crimenews_datetime, crimenews_locationLat, crimenews_locationLong) 
    VALUES 
        (?,?,?,?,?) 
    ON DUPLICATE KEY UPDATE 
        crimenews_url = VALUES( crimenews_url )';


$stmt = $conn->prepare( $sql );
$stmt->bind_param("sssss",$_POST['category'],$_POST['url'],$_POST['datetime'],$_POST['lat'],$_POST['lng']);
$stmt->execute();

您也许可以这样尝试-使用初始的select语句来确定url是否存在-未经测试

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

    session_start();

    try{

        $rowcount=0;

        $category = mysqli_real_escape_string($conn,$_POST['category'] );
        $url = mysqli_real_escape_string($conn,$_POST['url'] );
        $datetime = mysqli_real_escape_string($conn,$_POST['datetime'] );
        $lat = mysqli_real_escape_string($conn,$_POST['lat'] );
        $lng = mysqli_real_escape_string($conn,$_POST['lng'] );


        /*
            -------------------------------------
            does the url already exist in the db?
            use a SELECT statement to find out
            -------------------------------------
        */
        $sql='select `crimenews_url` from `crimenews` where `crimenews_url`=?';
        $stmt=$conn->prepare( $sql );
        if( !$stmt ) throw new Exception('Failed to prepare SELECT statement');

        $stmt->bind_param('s',$url );
        $result=$stmt->execute();

        if( $result ){
            $stmt->store_result();
            $rowcount=$stmt->num_rows;
            $stmt->free_result();
            $stmt->close();

        } else {
            throw new Exception('Initial SELECT query failed');
        }






        $sql='INSERT INTO `crimenews`
                ( `crimenews_cat`, `crimenews_url`, `crimenews_datetime`, `crimenews_locationLat`, `crimenews_locationLong` ) 
            VALUES 
                ( ?,?,?,?,? ) 
            ON DUPLICATE KEY UPDATE 
                crimenews_url = VALUES( `crimenews_url` )';


        $stmt = $conn->prepare( $sql );
        if( !$stmt )throw new Exception('Failed to prepare SQL insert statement');

        $stmt->bind_param( "sssss", $category, $url, $datetime, $lat, $lng );
        $result = $stmt->execute();


        /*
            ---------------------------------------------
            the following portion makes no sense and is 
            incorrect as no recordset will be returned by
            an `insert` statement.
            ----------------------------------------------
        */
        /*

        $result = $stmt->get_result();
        if( $result->num_rows > 0 ) {

            while( $row = $result->fetch_assoc() ){
                $crimenews_url = $row["crimenews_url"];
            }

            if( $crimenews_url == $url ){
                $_SESSION['message'] = "The news has added.";
                $_SESSION['msg_type'] = "success";
            } else {
                $_SESSION['message'] = "The news has already existed.";
                $_SESSION['msg_type'] = "danger";
            }
        }
        */


        /*
            ------------------------------------------------
            Instead, as there is now an initial SELECT query
            and we should have assigned a value to $rowcount
            you can use that for the logic test
            ------------------------------------------------
        */
        if( $rowcount==1 ){
            $_SESSION['message'] = "The news has already existed.";
            $_SESSION['msg_type'] = "danger";
        } else{
            $_SESSION['message'] = "The news has added.";
            $_SESSION['msg_type'] = "success";          
        }


        $stmt->close();
        header("location: main.php");


    }catch( Exception $e ){
        exit( $e->getMessage() );
    }
}

在php中使用准备好的语句后，会话消息未显示

1 个答案: