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执行此操作时，我不断收到此错误：

<?php
$info = $_POST['mname'];
$info = ucwords($info);

// Make a MySQL Connection
mysql_connect("localhost", "user", "password") or die(mysql_error());
mysql_select_db("javadatest") or die(mysql_error());  

// Get a specific result from the "example" table
$result = mysql_query("SELECT * FROM movieList
WHERE name = '$info'") or die(mysql_error());  

// get the first (and hopefully only) entry from the result
$row = mysql_fetch_array( $result );
// Print out the contents of each row into a table 
//echo $row['name']." ".$row['year']." ".$row['genre'];
echo "
  <script language=javascript>
  var jsvar;
  jsvar = <?php echo $row['name'], $row['year'], $row['genre'];?>
  function buy() {
    window.location = \"https://www.paypal.com\";   
    alert(\"Thanks for shopping at Movie Store\");     
  }
  var myWindow = window.open('', '', 'width = 300, height = 300);
  myWindow.document.write(jsvar);
  myWindow.document.write('<body>'); 

  myWindow.document.write('<input type="button" value="Buy" onclick=/"buy()/">');

  myWindow.document.write('</body>');

 //myWindow.buy = buy;
 </script>
";
?>

我试图通过将我的 javascript 代码放在我的 echo 语句中来在 php 文件中使用 javascript。我无法弄清楚我做错了什么。任何帮助，将不胜感激。

4

3 回答 3

1

你不需要写这一行

jsvar = <?php echo $row['name'], $row['year'], $row['genre'];?>

只需使用

jsvar = {$row['name']} {$row['year']} {$row['genre']};

因为你已经在 PHP 中了。

还

代替

myWindow.document.write('<input type="button" value="Buy" onclick=/"buy()/">');

和

myWindow.document.write('<input type=\"button\" value=\"Buy\" onclick=\"buy()\">');
于 2013-04-10T03:47:50.227 回答
0 
myWindow.document.write('<input type="button" value="Buy" onclick=/"buy()/">');

应该

myWindow.document.write(\"<input type='button' value='Buy' onclick='buy()'>\");
于 2013-04-10T03:48:25.060 回答
0 
    $row = mysql_fetch_array( $result );
    // Print out the contents of each row into a table 
    //echo $row['name']." ".$row['year']." ".$row['genre'];
    ?>
      <script language=javascript>
      var jsvar;
      jsvar = <?php echo json_encode($row['name']), json_encode($row['year']), json_encode($row['genre']);?>
      function buy() {
        window.location = \"https://www.paypal.com\";   
        alert(\"Thanks for shopping at Movie Store\");     
      }
...
</script>
于 2013-04-10T03:53:22.527 回答