数据没有发布到mySQL数据库

-3

我有以下的php脚本，但我插入到我的文本框中的数据没有被发布到数据库中，一旦按钮被点击。我没有收到错误，我不知道我的错误在哪里。数据没有发布到mySQL数据库

<?php 

if (isset($_POST['submit'])){ 
    $con = mysqli_connect("localhost","branchaw","211018556","basketball") or die("Could not connect"); 

    $pId = mysqli_real_escape_string($con,$_POST['pid']); 
    $Hometown = mysqli_real_escape_string($con,$_POST['Hometown']); 
    $Year = mysqli_real_escape_string($con,$_POST['Year']); 
    $Position = mysqli_real_escape_string($con,$_POST['Position']); 
    $Season = mysqli_real_escape_string($con,$_POST['Season']); 
    $Date_Of_Birth = mysqli_real_escape_string($con,$_POST['Date_Of_Birth']); 

    $query = "INSERT INTO `General` (`pId`,`Hometown`,`Year`,`Position`,`Season`,`Date_Of_Birth`) VALUES ('".$pId."','".$Hometown."','".$Year."','".$Position."''".$Season."','".$Date_Of_Birth."')"; 

    $result = mysqli_query($con,$query); 

    if($result){ 
    echo 'Data Inserted'; 
    } else{ 
    echo die(mysqli_error($con)); 
    } 

    //mysqli_free_result($result); 
    mysqli_close($con); 
} 
?> 

<html> 
    <head> 
    <title>Add Data to General Information</title> 
</head> 
<body>  
    <form method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>"> 
     <label>pId<input type="text" name="pid"/></label> 
     <label>Hometown<input type="text" name="Hometown"/></label> 
     <label>Year<input type="text" name="Year"/></label> 
     <label>Position<input type="text" name="Position"/></label> 
     <label>Season<input type="text" name="Season"/></label> 
     <label>Date of Birth<input type="text" name="Date_Of_Birth"/></label> 
    <br /> 
     <input type="submit" value="Add new data" name="submit"/>  
    </form> 
</body> 
</html>

来源

2016-12-14 Nate B

sqli_connect？使用mysqli_connect –

它仍然是不添加到我的数据库。出于某种原因，我没有收到数据插入或数据未插入的消息 –

参考http://www.tutorialrepublic.com/php-tutorial/php-mysql-insert-query.php –

<?php 

    if (isset($_POST['submit'])){ 
     $con = mysqli_connect("localhost","branchaw","211018556","basketball") or die("Could not connect"); 
     $pId = mysqli_real_escape_string($con,$_POST['pid']); 
     $query = "INSERT INTO `General` (`pId`) VALUES ('".$pId."')"; 

     $result = mysqli_query($con,$query); 

     if($result){ 
     echo 'Data Inserted'; 
     } else{ 
     echo 'Data Not Inserted'; 
     } 

     //mysqli_free_result($result); 
     mysqli_close($con); 
    } 
    ?> 

    <html> 
     <head> 
     <title>Add Data to General Information</title> 
    </head> 
    <body>  
     <form method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>"> 
      <label>pId<input type="text" name="pid"/></label> 
     <br /> 
      <input type="submit" value="Add new data" name="submit"/>  
     </form> 
    </body> 
    </html>

来源

2016-12-14 05:26:39

试试吧:) –

好吧，现在我收到数据未插入的错误消息 –

@NateB现在尝试echo'$ result'来查看您得到的SQL错误类型。 –

-1

$con = mysqli_connect("localhost","branchaw","211018556","basketball") or die("Could not connect"); 
    $pId = mysqli_real_escape_string($_POST['pId']); 
    $query = "INSERT INTO General (pId) VALUES ('$pId')" or die ("Query Failed");

没有必要在你的表名和字段名放'。

来源

2016-12-14 05:12:50

他没有在他的表字段和名称中放入一个''。这是一个倒退''，可以把它放在查询中。 –

确定不要把反拨“'在查询中，并尝试希望它会起作用。 –

数据没有发布到mySQL数据库

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