如何将此select sql语句更改为另一个select sql？

Question

我有代码下面的代码工作正常，但我想在这段代码中有一个sql语句而不是少数，所以当我尝试更改为一个我得到一个错误。

 $eventID = $_GET['id']; 

    $sql = "SELECT * FROM te_events where eventID='$eventID'"; 
    $result = $conn->query($sql); 

    while($row = $result->fetch_assoc()) 
    { 
     $eventTitle = $row['eventTitle']; 
     $eventDescription = $row['eventDescription']; 
     $eventStartDate = $row['eventStartDate']; 
     $eventEndDate = $row['eventEndDate']; 
     $eventPrice = $row['eventPrice']; 
     $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']; 
     } 
    } 
    ?> 

我改变了代码，但它似乎不工作。

<?php 
$eventID = $_GET['id']; 

    $sql = "SELECT * FROM te_events where eventID='$eventID' AND where venueID='$venueID' From te_venue AND where catID='$catID' From te_category"; 
    $queryresult = mysqli_query($conn, $sql) or die(mysqli_error($conn)); 
    while ($row = mysqli_fetch_array($queryresult)) { 
     $eventTitle = $row['eventTitle']; 
     $eventDescription = $row['eventDescription']; 
     $eventStartDate = $row['eventStartDate']; 
     $eventEndDate = $row['eventEndDate']; 
     $eventPrice = $row['eventPrice']; 
     $venueID = $row['venueID']; 
     $catID = $row['catID']; 
     $catName = $row['catDesc']; 
     $venueName = $row['venueName']; 



    } 

    ?> 

而我得到这个错误。

您的SQL语法错误;检查对应于你的MySQL服务器版本使用附近的正确语法手册“其中venueID =”“从te_venue并在CATID =‘’从te_category”在行1

Answer 1

是的，你可以这样做。但为了从三张表中获得所需的字段，您必须将它们结合在一起。

看看这篇关于W3S的文章：http://www.w3schools.com/sql/sql_join.asp。它解释了SQL JOIN语法和背后的基本理论。

如果你刚加入地点和事件SELECT语句的样子：

SELECT * FROM te_event 
JOIN te_venue 
ON te_vendue.venueID = te_event.venueID 
WHERE te_event.eventID = $eventID 

类别表是相似的。

注意：一般来说，使用SELECT *是不鼓励的。你应该列出你想从表格返回的字段。即。 SELECT te_eventID，te_venueID

Answer 2

您可以使用如下连接;

SELECT * FROM te_events 
JOIN te_venue ON te_events.venueID = te_venue.venueID 
JOIN te_category ON te_events.catID = te_category.catID 
WHERE eventID = :EventID 

如前所述，你也应该使用PDO's：

define("DB_DSN", "mysql:host=localhost;dbname=foo"); 
define("DB_USERNAME", "root"); 
define("DB_PASSWORD", "password"); 

// define sql 
$sSQL = "SELECT * FROM te_events 
JOIN te_venue ON te_events.venueID = te_venue.venueID 
JOIN te_category ON te_events.catID = te_category.catID 
WHERE eventID = :EventID"; 

// create an instance of the connection 
$conn = new PDO(DB_DSN, DB_USERNAME, DB_PASSWORD); 

// prepare 
$st = $conn->prepare($sSQL); 

// securely bind any user input in the query 
$st->bindValue(":EventID", $iEventID, PDO::PARAM_INT); 

// execute the connection 
$st->execute() 

$aResults = array(); 

// loop over results and store in aResults 
while($row = $st->fetch()){ 

    $aResults[] = $row; 

} 

// output data 
foreach($aResults as $aResult){ 
    echo "title: ".$aResult['eventTitle']; 
} 

你应该总是避免使用select *。特别是当你加入时。如果你需要重复的列名，确保你只请求你需要的和别名