Question

我在将数据插入表格时遇到问题。我有以下表格：track（id，tracktitle），album（id，albumtitle），composer（id，composername），albumtrack（PRIMARY：trackid，albumid，composerid）。

我的PHP页面允许您添加曲目，然后选择与其相关的专辑和作曲家。它可以将曲目添加到曲目表中，但不会将其添加到专辑中。

我一直在四处寻找如何去做，而且我一点都不知所措。任何人都可以告诉我应该如何正确地做到这一点？感谢

if (isset($_POST['tracktitle'])): 
  // A new track has been entered
  // using the form.

  $cid= $_POST['cid'];
  $tracktitle = $_POST['tracktitle'];
  $albs = $_POST['albs'];

  if ($cid == '') {
  exit('<p>You must choose an composer for this track. Click "Back" and try         again.</p>');   }

  $sql = "INSERT INTO track, albumtrack SET
  track.tracktitle='$tracktitle', albumtrack.albumid='$albs',    albumtrack.composerid='$cid' " ;
  if (@mysql_query($sql)) {
  echo '<p>New track added</p>';
  } else {
  exit('<p>Error adding new track' . mysql_error() . '</p>');
  }

  $trackid = mysql_insert_id();


 if (isset($_POST['albs'])) {
 $albs = $_POST['albs'];
  } else {
 $albs = array();
 }

 $numAlbs = 0;
  foreach ($albs as $albID) {
 $sql = "INSERT IGNORE INTO albumtrack
        SET albumtrack.trackid='$trackid', albumtrack.albumid='$albs',     albumtrack.composerid='$cid'";
if ($ok) {
  $numAlbs = $numAlbs + 1;
} else {
  echo "<p>Error inserting track into album $albID: " .
      mysql_error() . '</p>';
}
  }
 ?>

 <p>Track was added to <?php echo $numAlbs; ?> albums.</p>

  <p><a href="<?php echo $_SERVER['PHP_SELF']; ?>">Add another track</a></p>
  <p><a href="tracks.php">Return to track search</a></p>

 <?php
 else: // Allow the user to enter a new track

  $composers = @mysql_query('SELECT id, composername FROM composer');
   if (!$composers) {
exit('<p>Unable to obtain composer list from the database.</p>');
   }

  $albs = @mysql_query('SELECT id, albumtitle FROM album');
 if (!$albs) {
  exit('<p>Unable to obtain album list from the database.</p>');
  }
  ?>

  <form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post">
  <p>Enter the new track:<br />
  <textarea name="tracktitle" rows="1" cols="20">
   </textarea></p>
   <p>Composer:
   <select name="cid" size="1">
  <option selected value="">Select One</option>
  <option value="">---------</option>
    <?php
  while ($composer= mysql_fetch_array($composers)) {
   $cid = $composer['id'];
   $cname = htmlspecialchars($composer['composername']);
   echo "<option value='$cid'>$cname</option>\n";
  }
?>
 </select></p>
 <p>Place in albums:<br />
  <?php
   while ($alb = mysql_fetch_array($albs)) {
   $aid = $alb['id'];
   $aname = htmlspecialchars($alb['albumtitle']);
   echo "<label><input type='checkbox' name='albs[]'      value='$aid' />$aname</label><br />\n";
  }
  ?>

Answer 1

您的插入句子错误：

试试这个：

$sql = "INSERT IGNORE INTO albumtrack (trackid, albumid, composerid) values " .
       "($trackid, $albs, $cid)";

假设您的 ID 是数字。

当您注入从请求中获得的非清理值时，请注意SQL注入。

Answer 2

除了Pablo更正的编写查询的方式。我还注意到你试图一次插入两个表：

$sql= "INSERT INTO track, albumtrack"

MySQL不允许一次插入两个表。

将数据插入SQL表

2 个答案: