mysql中的列名無效PHP查詢

我有以下表單正常工作並填充MySQL數據庫。自從我在數據庫中添加了一個新字段「類型」並將其添加到表單中後，不過，現在當我嘗試添加一個新條目時，它會顯示「無效的列名」鍵入'「。任何幫助將不勝感激。mysql中的列名無效PHP查詢

<script> 
     $(document).ready(function() { 
     $("#datepicker").datepicker(); 
     }); 
     </script> 

    <script> 
$(document).ready(function() { 
$("#datepicker2").datepicker(); 
}); 
</script> 

</head> 
<body> 
<?php 
// if there are any errors, display them 
if ($error != '') 
{ 
echo '<div style="padding:4px; border:1px solid red; color:red;">'.$error.'</div>'; 
} 
?> 

    <form action="" method="post"> 
<div> 
<label for="Posted"><strong>Posted: </strong> </label> 
<input id="datepicker" name="posted" value="<?php echo $Posted; ?>" /><br/><br/> 

<label for="Ends"><strong>Ends: </strong> </label> 
<input id="datepicker2" name="ends" value="<?php echo $Ends; ?>" /><br/><br/> 

<label for="Position"><strong>Position: </strong></label> 
    <input type="text" name="position" value="<?php echo $Position; ?>" /><br/><br/> 

    <label for="Location"><strong>Location: </strong> </label> 
<select name="location"> 
    <option value=" ">Select...<?php echo $Location; ?></option> 
    <option value="Fargo">Fargo</option> 
    <option value="Grand Forks">Grand Forks</option> 
</select><br/><br/> 

<label for="Application Type"><strong>Application Type: </strong> </label> 
    <select name="type"> 
    <option value=" ">Select...<?php echo $Type; ?></option> 
    <option value="Driver">Driver</option> 
    <option value="Employee">Employee</option> 
</select><br/><br/> 

<label for="Hours"><strong>Hours: </strong> </label> 
<input type="text" name="hours" value="<?php echo $Hours; ?>" /><br/><br/> 

<label for="Pay"><strong>Pay: </strong> </label> 
<input type="text" name="pay" value="<?php echo $Pay; ?>" /><br/><br/> 

<label for="Benefits"><strong>Benefits: </strong> </label> 
<textarea cols="60" rows="2" name="benefits" value="<?php echo $Benefits; ?>" ><?php echo $Benefits; ?></textarea><br/><br/> 

<label for="Description"><strong>Description: </strong> </label> 
<textarea cols="60" rows="3" name="description" value="<?php echo $Description; ?>" ><?php echo $Description; ?></textarea><br/><br/> 


<input type="submit" name="submit" value="Submit"> 
</div> 
</form> 
</body> 
</html> 
<?php 
} 




// connect to the database 
include('connect-db.php'); 

// check if the form has been submitted. If it has, start to process the form and save it to the database 
if (isset($_POST['submit'])) 
{ 
// get form data, making sure it is valid 

function ms_escape_string($data) { 
     if (!isset($data) or empty($data)) return ''; 
     if (is_numeric($data)) return $data; 

     $non_displayables = array(
      '/%0[0-8bcef]/',   // url encoded 00-08, 11, 12, 14, 15 
      '/%1[0-9a-f]/',    // url encoded 16-31 
      '/[\x00-\x08]/',   // 00-08 
      '/\x0b/',     // 11 
      '/\x0c/',     // 12 
      '/[\x0e-\x1f]/'    // 14-31 
     ); 
     foreach ($non_displayables as $regex) 
      $data = preg_replace($regex, '', $data); 
     $data = str_replace("'", "''", $data); 
     return $data; 
    } 

    ms_escape_string($_POST); 


    $posted=$_POST['posted']; 
    $ends=$_POST['ends']; 
    $type=$_POST['type']; 
    $position=$_POST['position']; 
    $location=$_POST['location']; 
    $hours=$_POST['hours']; 
    $pay=$_POST['pay']; 
    $benefits=$_POST['benefits']; 
    $description=$_POST['description']; 

// check to make sure all fields are entered 
if ($posted == '' || $ends == '' || $type == '' || $position == '' || $location == '' || $hours == '' || $pay == '' || $benefits == '' || $description == '') 
{ 
// generate error message 
$error = 'ERROR: Please fill in all required fields!'; 

// if any fields are blank, display the form again 
renderForm($posted, $ends, $type, $position, $location, $hours, $pay, $benefits, $description, $error); 
} 
else 
{ 
// save the data to the database 
    $SQL = "INSERT INTO JobPosting (posted, ends, type, position, location, hours, pay, benefits, description) VALUES ('$posted', '$ends', '$type', '$position', '$location', '$hours', '$pay', '$benefits', '$description')"; 

    $result = mssql_query($SQL) 
     or die (mssql_get_last_message()); 

// once saved, redirect back to the view page 
header("Location: view.php"); 
} 
} 
else 
// if the form hasn't been submitted, display the form 
{ 
renderForm('','','','','','','','','',''); 
}

來源

2012-10-23 Djacksway

'type'是MySQL中的保留字。理想情況下，選擇另一個列名稱。 –

在MySQL中鍵入保留字？如果是這樣，您將需要在查詢中將其轉義。即'\'輸入\'' – thatidiotguy

爲什麼MySQL在標題和標籤中？你的代碼顯示你使用MsSQL。 – Pitchinnate

您需要在查詢即type逃脫列名type因爲type是在MySQL的語法時才保留工作。

你到底怎麼逃避？

來源

2012-10-23 20:47:50 thatidiotguy

非常感謝。我以爲我瘋了一秒。非常感激。 – Djacksway

type是MS SQL中的保留字。這意味着它在其他MS SQL操作中使用。同樣，你會遇到麻煩，如果你用括號有一個名爲select或group

可以逃脫保留字柱：[]

$SQL = "INSERT INTO JobPosting (posted, ends, [type], position, location, hours, pay, benefits, description) VALUES ('$posted', '$ends', '$type', '$position', '$location', '$hours', '$pay', '$benefits', '$description')";

來源

2012-10-23 20:50:23 Chris

mysql中的列名無效PHP查詢

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