[英]What is wrong with my SQL Insert code?
我正在努力找出为什么这段代码对我不起作用。 我有表格: albums (albumid, albumname)
, composers (composerid, composername)
和tracks (trackid, tracktitle, albumid, composerid)
。
当我使用表单添加曲目并将其链接到作曲家和专辑时:
<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['composerid'];
$cname = htmlspecialchars($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['albumid'];
$aname = htmlspecialchars($alb['albumname']);
echo "<label><input type='checkbox' name='albs[]'
value='$aid' />$aname</label><br />\n";
} ?>
</p>
<input type="submit" value="SUBMIT" />
</form>
<?php endif; ?>
我收到此消息:
添加了新曲目
将曲目插入专辑2时出错:
曲目已添加到0个相册。
表单之前的php代码为:
if (isset($_POST['tracktitle'])):
// A new track has been entered
// using the form.
$tracktitle = mysql_real_escape_string($tracktitle);
$cid= $_POST['cid'];
$tracktitle = $_POST['tracktitle'];
$albs = $_POST['albs'];
if ($cid == '') {
exit('<p>You must choose an composer for this track. Click
“返回”,然后重试。
');} $sql = "INSERT INTO tracks (tracktitle) VALUES ('$tracktitle')" ; if (@mysql_query($sql)) { echo '<p>New track added</p>'; } else { exit('<p>Error adding new track' . mysql_error() . '</p> echo mysql_error() ');} $trackid = mysql_insert_id(); if (isset($_POST['albs'])) { $albs = $_POST['albs']; } else { $albs = array(); } $numAlbs = 0; foreach ($albs as $albID) { $sql = "INSERT IGNORE INTO tracks (trackid, albumid, composerid) VALUES " . "($trackid, $albs, $cid)"; if ($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> <?php else: // Allow the user to enter a new track $composers = @mysql_query('SELECT composerid, composername FROM composers'); if (!$composers) { exit('<p>Unable to obtain composer list from the database.</p>'); } $albs = @mysql_query('SELECT albumid, albumname FROM albums'); if (!$albs) { exit('<p>Unable to obtain album list from the database.</p>');}?>
我一直在寻找失败的原因,而且还在不断撞墙。 我还知道,目前还不是很安全,这是我接下来要解决的问题。 我只想让实际功能先工作。
@paj:更改
if ($ok) {
至
if (mysql_query($sql)) {
--
我还建议您将SQL语句更新为
$sql = "INSERT INTO tracks (tracktitle) VALUES ('" . $tracktitle . "')";
$sql = "INSERT IGNORE INTO tracks (trackid, albumid, composerid) VALUES (" . $trackid . ", " . $albID . ", " . $cid . ")";
在我看来, $ok
if ($ok) {
不存在,除非在if ($ok) {
行中。 需要在某个地方定义它,否则它将始终显示为false,因为它不存在。
实际上,您可以跳过不存在的$ ok,并在if (@mysql_query($sql)) {
插入上面的代码。 我确实必须同意注释,即代码需要一些爱,但是如果您想知道为什么它崩溃了,看来这就是原因。
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