Question

我有代码，下面的代码工作正常，但我希望在此代码中有一个sql语句而不是少数，所以当我尝试更改为一个时，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'];
      }
    }
  ?>

我将代码更改为此，但似乎无效。

 <?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'];



    }

  ?>

我得到了这个错误。

您的SQL语法有错误;检查与您的MySQL服务器版本相对应的手册，以便在“场地ID =＆＃39;＆＃39;”附近使用正确的语法。来自te_venue和catID =＆＃39;＆＃39;来自te_category＆＃39;在第1行

Answer 1

是的，你可以这样做。但是为了从所有三个表中获取所需的字段，您必须将它们连接在一起。

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

如果您刚加入场地和活动您的选择语句如下：

from tkinter import StringVar
class StringVar_WithHistory(StringVar):
    def __init__(self):
        self.history = []

    def set (self,value):
        if StringVar.get(self) != value:
            StringVar.set(self, value)
            self.history.append(value)

    def undo (self):
        StringVar.set(self,history[-1])
        self.history.pop()

from tkinter import OptionMenu
class OptionMenuUndo(OptionMenu):
    def __init__(self,parent,title,*option_tuple,**configs):
        self.result = StringVar_WithHistory()
        self.result.set(title)
        OptionMenu.__init__(self,parent,self.result,*option_tuple,**configs)


    def get(self):                
        return self.result.get()


    def undo(self):
        self.result.undo()     


    def simulate_selection(self,option):
        self.result.set(option)

类别表类似。

注意：一般情况下，不建议使用 SELECT * 。您应列出要从表中返回的字段。即。 SELECT te_eventID，te_venueID

Answer 2

您可以使用以下联接;

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

如上所述，您还应该使用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'];
}

您应该始终避免使用select *。特别是当你加入时。如果需要重复的列名，请确保仅请求所需内容和别名

如何将这个select sql语句更改为另一个select sql？

2 个答案: