【发布时间】:2016-11-15 00:40:16
【问题描述】:
这是我目前工作正常的代码:
<?php
$sql = "SELECT * FROM te_events order by eventTitle ASC ";
$result = $conn->query($sql);
while($row = $result->fetch_assoc())
{
$venueID = $row['venueID'];
$catID = $row['catID'];
$sql2 = "SELECT * FROM te_venue where venueID='$venueID'";
$result2 = $conn->query($sql2);
while($row2 = $result2->fetch_assoc())
{
$venueName = $row2['venueName'];
}
$sql3 = "SELECT * FROM te_category where catID='$catID'";
$result3 = $conn->query($sql3);
while($row3 = $result3->fetch_assoc())
{
$catName = $row3['catDesc'];
}
?>
但我想把它改成这种格式。我只能做到这一点,直到我得到错误为止。
<?php
$sql ="SELECT eventTitle, eventID, venueID, catID, eventStartDate, eventEndDate, eventPrice FROM te_events ORDER BY eventTitle ASC";
$queryresult = mysqli_query($conn, $sql) or die(mysqli_error($conn));
while ($row = mysqli_fetch_array($queryresult)) {
$venueID = $row['venueID'];
$catID = $row['catID'];
$venueName = $row['venueName'];
$catName = $row['catDesc'];
?>
那我该怎么做呢?
如何连接两个表?
【问题讨论】:
-
那我该怎么做呢?
-
如何连接两个表?
-
没有代码的问题对未来的访问者没有意义。请保留问题中的代码。如果有一些安全数据(我没有看到),你应该在你的服务器上重置它,因为它已经暴露了。