【发布时间】:2020-02-26 23:49:16
【问题描述】:
我似乎无法让每部电影都出现在各自的日期下,并且为电影时间选择提供正确的时间下拉菜单
$sql_date = "SELECT DISTINCT sDate, sTitle FROM movieScreenings ";
$result_date = mysqli_query($db, $sql_date);
while($row = mysqli_fetch_array($result_date)) {
echo "<h2>" . $row['sDate'] . "</h2>";
$sql_movie = "SELECT * FROM movieList, movieScreenings WHERE title = '" . $row['sTitle'] . "'";
$result_movie = mysqli_query($db, $sql_movie);
while($row2 = mysqli_fetch_array($result_movie)) {
echo "<div class='box'>
<img class='poster' src='posters/" . $row2['poster'] . "'/>
<h2>" . $row2['title'] . "</h2>
<p>" . $row2['description'] . "</p>";
$sql_time = "SELECT DISTINCT sTime FROM movieScreenings WHERE sTitle = '" . $row2['title'] . "' AND sDate = '" . $row['sDate'] . "'";
$result_time = mysqli_query($db, $sql_time);
while($row3 = mysqli_fetch_array($result_time)) {
echo "<select name='sTime'>
<option value='" . $row3['sTime'] . "'>" . $row3['sTime'] . "</option>
</select>";
echo "</div>";
}
}
}
mysqli_free_result($result_date);
mysqli_free_result($result_movie);
mysqli_free_result($result_time);
// Close connection
mysqli_close($db);
【问题讨论】:
-
您的代码易受 SQL 注入攻击。您应该使用准备好的语句。
-
嗨,Kelvin,我们需要更多关于您正在使用的数据以及您想要实现的目标的信息......
标签: php html mysql sql phpmyadmin