Question

//这个PHP代码试图在数据库表中插入一些值＆＃34; transaction_db＆＃34;。插入查询与要插入的数据一起运行。

// HTML表单用于测试PHP代码。但是在提交HTML表单后，没有显示新插入的行，也没有任何错误。

//下面的PHP代码

<?PHP
include_once("connection.php");

session_start();

if(isset($_POST['event_date']) && isset($_POST['event_location']) && isset ($_POST['organisation_name']) && isset ($_POST['client_name']) && isset ($_POST['client_type']) )    
{

$client_name= mysqli_real_escape_string($con,$_POST['client_name']);
$client_type= mysqli_real_escape_string($con,$_POST['client_type']);
$event_date= mysqli_real_escape_string($con,$_POST['event_date']); 
$organisation_name= mysqli_real_escape_string($con,$_POST['organisation_name']);
$event_location= mysqli_real_escape_string($con,$_POST['event_location']);

$score1= mysqli_real_escape_string($con,$_POST['score1']);
$score2= mysqli_real_escape_string($con,$_POST['score2']);
$score3= mysqli_real_escape_string($con,$_POST['score3']);
$score4= mysqli_real_escape_string($con,$_POST['score4']);
$score5= mysqli_real_escape_string($con,$_POST['score5']);
$score6= mysqli_real_escape_string($con,$_POST['score6']);
$score7= mysqli_real_escape_string($con,$_POST['score7']);
$score8= mysqli_real_escape_string($con,$_POST['score8']);
$score9= mysqli_real_escape_string($con,$_POST['score9']);
$score10= mysqli_real_escape_string($con,$_POST['score10']);
$answer1= mysqli_real_escape_string($con,$_POST['answer1']);
$answer2= mysqli_real_escape_string($con,$_POST['answer2']);
$answer3= mysqli_real_escape_string($con,$_POST['answer3']);
$answer4= mysqli_real_escape_string($con,$_POST['answer4']);
$answer5= mysqli_real_escape_string($con,$_POST['answer5']);
$comments= mysqli_real_escape_string($con,$_POST['comments']);

$login_id= mysqli_real_escape_string($con,$_POST['login_id']);
$lid = (int)$login_id;


$query = "INSERT INTO transaction_db (client_name,client_type,event_date,organisation_name,event_location,,login_id,score1,score2,score3,score4,score5,score6,score7,score8,score9,score10,answer1,answer2,answer3,answer4,answer5,comments) VALUES ('$client_name','$client_type','$event_date','$organisation_name','$event_location','$login_id','$score1','$score2','$score3','$score4','$score5','$score6','$score7','$score8','$score9','$score10','$answer1','$answer2','$answer3','$answer4','$answer5','$comments')";

if(mysqli_query($con,$query)){
     $data['success'] = 1;
     echo json_encode($data);
     }
    else{
        $data['success'] = 0;
    echo json_encode($data);
    }
}
mysqli_close($con);
?>

// HTML FORM CREATION

<html>

   <head>
      <title>client welcome Page</title>

      <style type = "text/css">
         body {
            font-family:Arial, Helvetica, sans-serif;
            font-size:14px;
         }

         label {
            font-weight:bold;
            width:100px;
            font-size:14px;
         }

         .box {
            border:#666666 solid 1px;
         }
      </style>

   </head>

   <body bgcolor = "#FFFFFF">

      <div align = "center">
         <div style = "width:300px; border: solid 1px #333333; " align = "left">
            <div style = "background-color:#333333; color:#FFFFFF; padding:3px;"><b>Login</b></div>

            <div style = "margin:30px">

               <form action = "" method = "post">
                  <label>name  :</label><input type = "text" name = "name" class = "box"/><br /><br />
                  <label>type  :</label><input type = "text" name = "type" class = "box" /><br/><br />
                  <label>date  :</label><input type = "text" name = "date" class = "box" /><br/><br />
                  <label>org name  :</label><input type = "text" name = "org name" class = "box" /><br/><br />
                  <label>location  :</label><input type = "text" name = "location" class = "box" /><br/><br />
                  <input type = "submit" value = " Submit "/><br />
               </form>

               <div style = "font-size:11px; color:#cc0000; margin-top:10px"></div>

            </div>

         </div>

      </div>

   </body>
</html>

Answer 1

你应该使用盲目查询来保证安全，但既然你没有使用它，那么试着改用这个mysql。调试它的错误位置更容易。

INSERT INTO transaction_db SET 
client_name = '$client_name'
,client_type = '$client_type'
,event_date = '$event_date'
,organisation_name = '$organisation_name'
,event_location = '$event_location'
,login_id = '$login_id'
,score1 = '$score1'
,score2 = '$score2'
,score3 = '$score3'
,score4 = '$score4'
,score5 = '$score5'
,score6 = '$score6'
,score7 = '$score7'
,score8 = '$score8'
,score9 = '$score9'
,score10 ='$score10'
,answer1 = '$answer1'
,answer2 = '$answer2'
,answer3 = '$answer3'
,answer4 = '$answer4'
,answer5 = '$answer5'
,comments = '$comments'

数据无法插入数据库表中

1 个答案: