多個查詢MySQL PHP

是否有正確的方法來執行此操作。我想計算一個表的平均評分，並同時在另一個表中更新結果。林新的PHP和MYSQL和我將不勝感激任何幫助多個查詢MySQL PHP

$query=mysql_query("INSERT INTO review (username, restaurant, rating, review) VALUES ('$username','$restaurant','$rating','$review')"); 
if($query) 
{ 
    $avg_query="SELECT ROUND(AVG(rating),0) FROM review WHERE name =\"$restaurant\""; 
    $avg_result=mysql_query($avg_query); 
    $avg_row=mysql_fetch_array($avg_result); 
    $rating=$row['ROUND(AVG(rating),0)']; 
    if($avg_result) 
    { 
     $update_query= "UPDATE restaurant SET rating=\"$rating\" WHERE name =\"$restaurant\""; 
     $update_result=mysql_query($update_query); 
    } 
} 
else 
{ 
}

謝謝！

來源

2012-08-04 JProg

聽起來像是你想要一個子查詢 – 2012-08-04 06:28:22

使用參數化查詢！切勿將變量直接插入查詢字符串！ http://stackoverflow.com/questions/60174/best-way-to-prevent-sql-injection-in-php – 2012-08-04 06:38:31

UPDATE restaurant 
SET rating= (SELECT ROUND(AVG(rating),0) FROM review WHERE name ='$restaurant') 
WHERE name ='$restaurant'

來源

2012-08-04 06:29:58

我將兩者結合成一個像這樣的：

$query=mysql_query("INSERT INTO review (username, restaurant, rating, review) VALUES ('$username','$restaurant','$rating','$review')"); 
if($query) 
{ 
    $avg_query="UPDATE restaurant a SET rating=(SELECT ROUND(AVG(rating),0) FROM review WHERE name =a.name) WHERE name ='".$restaurant."'"; 
    $avg_result=mysql_query($avg_query); 
} 
else 
{ 
}

話雖如此，由於mysql_*功能貶值，你應該動過要麼PDO或mysqli的。

來源

2012-08-04 06:31:23 Fluffeh

另一種選擇是使用mysql trigger。例如（不要抱着我的語法）：

CREATE TRIGGER after_insert_review 
AFTER INSERT ON review 
FOR EACH ROW 
BEGIN 
    UPDATE restaurant 
    SET rating = (SELECT ROUND(AVG(rating),0) FROM review WHERE name = NEW.restaurant) 
    WHERE name = NEW.restaurant; 
END

再次如其他人所說

使用PDO或MySQLi。

來源

2012-08-04 06:47:23 Kurt

多個查詢MySQL PHP

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