將多個表中的數據插入到一個單獨的表中

首先，我知道MySQL是一種折舊語言，但我的主管堅持使用它。將多個表中的數據插入到一個單獨的表中

我在我的數據庫中有三個表。他們是：「學生」，「導師」，「匹配」。當我找到運行下面的查詢（$ query2）時，我正在檢查確保該查詢中的所有條件都完全匹配。

當完美匹配我想插入以下到匹配表：

"insert into match (tutor_id, student_id, school_id) values ('$tutor_id','$student_id','$school_id')";

的「匹配」表中包含列：

match_id(auto incremented), tutor_id, student_id, school_id

任何幫助將不勝感激。

if(mysql_query($query2)){ 
    $check_availability = "select * 
          from tutors, students 
          where tutor_availability = '$student_availability' 
          AND (tutor_subject_1 = '$student_subject_1' 
          OR tutor_subject_1 = '$student_subject_2' 
          OR tutor_subject_2 = '$student_subject_2' 
          OR tutor_subject_2 = '$student_subject_1')";  

    $run_5 = mysql_query($check_availability); 

    if(mysql_num_rows($run_5)>0) { 

     echo "<script>alert('we have a match')</script>"; 
     my_sql_query($query3) = "insert into match (tutor_id, student_id, school_id) values ('$tutor_id','$student_id','$school_id')"; 

     mysql_query($query3); 
    } else{ 
     echo "<script>alert('no match found')</script>"; 
    } 
}

來源

2014-03-26 bano590

MySQL是絕不棄用的語言，PHP mysql_ *功能 –

那麼什麼是您的問題...？ – user1844933

這些值不會進入「匹配」表 – bano590

我覺得這條線是錯誤的：

my_sql_query($query3) = "insert into match (tutor_id, student_id, school_id) values ('$tutor_id','$student_id','$school_id')";

函數名稱拼寫錯誤，參數不應該是$query3但實際查詢：

mysql_query("insert into match (tutor_id, student_id, school_id) values ('$tutor_id','$student_id','$school_id')");

來源

2014-03-26 10:24:06 Typoheads

對不起，這是一個小錯誤。更正了它，但它仍然不起作用 – bano590

基本上你將前一個查詢的輸出插入到數據庫中。

變化

$check_availability = "select * 
         from tutors, students 
         where tutor_availability = '$student_availability' 
         AND (tutor_subject_1 = '$student_subject_1' 
         OR tutor_subject_1 = '$student_subject_2' 
         OR tutor_subject_2 = '$student_subject_2' 
         OR tutor_subject_2 = '$student_subject_1')";

到

$check_availability = "insert into match (tutor_id, student_id, school_id) 
         select tutors.id, students.id, students.school_id 
         from tutors, students 
         where tutor_availability = '$student_availability' 
         AND (tutor_subject_1 = '$student_subject_1' 
         OR tutor_subject_1 = '$student_subject_2' 
         OR tutor_subject_2 = '$student_subject_2' 
         OR tutor_subject_2 = '$student_subject_1')";

來源

2014-03-26 10:25:27

有了這段代碼，由於某種原因它不再找到匹配項？我認爲它肯定會工作 – bano590

@ bano590 - 這是一個插入查詢，而不是一個選擇。如果你想檢查匹配，那麼你將不得不運行你的選擇，然後遍歷行，然後插入。 –

這種情況下最好的方法是什麼？我對MySQL – bano590

您可以通過該行必須循環得到插入值：

if(mysql_query($query2)){ 
    $check_availability = "select * 
          from tutors, students 
          where tutor_availability = '$student_availability' 
          AND (tutor_subject_1 = '$student_subject_1' 
          OR tutor_subject_1 = '$student_subject_2' 
          OR tutor_subject_2 = '$student_subject_2' 
          OR tutor_subject_2 = '$student_subject_1')";  

    $run_5 = mysql_query($check_availability); 

    if(mysql_num_rows($run_5)>0) { 

     echo "<script>alert('we have a match')</script>"; 
     while($row = mysql_fetch_assoc($run_5)) { 
      $tutor_id = $row['tutor_id']; 
      $student_id = $row['student_id']; 
      $school_id = $row['school_id']; 
      mysql_query("insert into match (tutor_id, student_id, school_id) values ('$tutor_id','$student_id','$school_id')"); 
     } 

    } else{ 
     echo "<script>alert('no match found')</script>"; 
    } 
}

我不知道確切的字段名，但它應該指向正確的方向。

來源

2014-03-26 10:41:18 Typoheads

相當陌生在這種情況下查詢$ res是什麼？ – bano590

@ bano590 $ res應該是$ run_5。對不起，我在示例代碼中糾正了這個問題。 – Typoheads

你實際上得到了所有的字段名稱，但它仍然不起作用 – bano590

也許嘗試使用mysql_fetch_row（$ run_5）並將選擇查詢更改爲只拉出tutor_id，student_id和school_id。

if(mysql_query($query2)){ 
$check_availability = "select t.tutor_id, s.student_id, t.school_id 
         from t.tutors, s.students 
         where t.tutor_availability = '$student_availability' 
         AND (t.tutor_subject_1 = '$student_subject_1' 
         OR t.tutor_subject_1 = '$student_subject_2' 
         OR t.tutor_subject_2 = '$student_subject_2' 
         OR t.tutor_subject_2 = '$student_subject_1')";  

$run_5 = mysql_query($check_availability); 

if(mysql_num_rows($run_5)>0) { 

    echo "<script>alert('we have a match')</script>"; 
    while($row = mysql_fetch_row($run_5)) { 
     $tutor_id = $row[0]; 
     $student_id = $row[1]; 
     $school_id = $row[2]; 
     mysql_query("insert into match (tutor_id, student_id, school_id) values ('$tutor_id','$student_id','$school_id')"); 
    } 

} else{ 
    echo "<script>alert('no match found')</script>"; 
}

}

來源

2014-03-26 16:25:39 zombie20

我試過這段代碼，它是最接近我想要的答案。問題似乎是這樣：mysql_query（「insert into match（tutor_id，student_id，school_id）values（'$ tutor_id'，'$ student_id'，'$ school_id'）」）; – bano590

將多個表中的數據插入到一個單獨的表中

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