-4

我是新手

下面是我的 HTML PHP 表单

<div id="user">
<form name="booksInput" action="theGamer.php" method="post">
<p>This Section is to add new USer Record in DB</p>
User F Name: <input type="text" name="Fuser">
User L Name: <input type="text" name="Luser">
<?php
$con=mysql_connect("localhost","root","","library");
mysql_select_db("library",$con);
$sqlb = "SELECT * FROM books";
$queryb = mysql_query($sqlb);
while ($resultsb[] = mysql_fetch_object ( $queryb ));
?>
Book Name:<select name="bID">
<?php foreach ( $resultsb as $optionb ) { ?>
<option value="<?php echo $optionb->bID; ?>"><?php echo $optionb->book;?></option>
<?php } ?>
</select><br>
<?php               
$con=mysql_connect("localhost","root","","library");
mysql_select_db("library",$con);
$sqlc = "SELECT * FROM country";
$queryc = mysql_query($sqlc);
while ( $resultsc[] = mysql_fetch_object ( $queryc ) );
array_pop($resultsc);
?>
Country Name:<select name="cID">
<?php foreach ( $resultsc as $optionc ) : ?>
<option value="<?php echo $optionc->cID; ?>"><?php echo $optionc->country;?>/option

    <?php endforeach; ?>
</select>
<input type="submit" value="Submit" name="subUser">
</form> 
    </div>

现在，当提交按钮名称sunUser被按下时theGamer.php，代码theGamer.php如下

<?php
$con=mysql_connect("localhost","root","","library");
function insertBook()
{
global $con;
mysql_select_db("library",$con);
$book=mysql_real_escape_string($_POST['books']);
$sql="INSERT INTO books (bID, book) VALUES ('','$book')";
if (!mysql_query($sql,$con))
{
die('Error: ' . mysql_error($con));
}`
echo "1 record added";
}

function insertCountry()
{
global $con;
mysql_select_db("library",$con);
$country=mysql_real_escape_string($_POST['country']);
$sql="INSERT into country (cID, country) VALUES ('','$country')";

if (mysql_query($sql,$con))
{
echo "1 record added";
}
else{die('Error: ' . mysql_error($con));}

}

function insertUser()
{
$con=mysql_connect("localhost","root","","library");
mysql_select_db("library",$con);
$Fuser=mysql_real_escape_string($_POST['Fuser']);
$Luser=mysql_real_escape_string($_POST['Luser']);
$bID=mysql_real_escape_string($_POST['bID']);
$cID=mysql_real_escape_string($_POST['cID']);
$sql="INSERT INTO user (userid, uFname,uLname, bID, CID ) VALUES  
    ('','$Fuser','$Luser','bID','cID')";
if (!mysql_query($sql,$con))
{
die('Error: ' . mysql_error($con));
}
echo "1 record added";
    }

   //Code for Knowing which button is pressed
  if(isset($_POST['subBooks']))
  {
  insertBook();
  } 
 if(isset($_POST['subCountry']))
 {
 insertCountry();
 }
if(isset($_POST['subUser']))
 {
 insertUser();
 } 
?>

它给出了这个错误 Parse error: parse error, Expecting T_STRINGor T_VARIABLEor T_NUM_STRINGinC:\wamp\www\library\theGamer.php在第 20 行，而第 20 行有 np 关注，因为subUser按钮被按下并且 if 语句必须u=insertuser运行，为什么它在第 20 行给出错误

问候阿普

4

1 回答 1

0 
$book=mysql_real_escape_string($_POST['books']);
$sql="INSERT INTO books (bID, book) VALUES ('','$book')";
if (!mysql_query($sql,$con))
{
die('Error: ' . mysql_error($con));
}`   <---  Here is a rogue quote..

如果 StackOverflow 发现了这个问题，语法高亮。你也使用带有荧光笔的编辑器吗？

由于代码现在无效，PHP 无法解析文件并将失败。你实际上不执行这段代码并不重要，因为整个文件都被解析了。这与运行时错误不同，运行时错误发生在代码解析之后和代码执行时。在您的情况下，错误发生在任何 PHP 代码执行之前。

于 2013-07-28T07:57:08.523 回答