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我有3個表。 tbl_info,tbl_trainings和tbl_infotrainings,其中tbl_infotrainings用於連接tbl_info和tbl_trainings。已加入表INSERT語句
tbl_trainings接受多個複選框值。 tbl_trainings包含:
+------+-----------+
| id | training |
+------+-----------+
tbl_infotrainings包含:
+------+-----------+---------------+
| id | info_id | training_id |
+------+-----------+---------------+
我在這裏的問題是,插入新的信息時。這是我使用的代碼:用於插入tbl_info的id
(正常工作)
$sql3 = "INSERT INTO tbl_infotrainings (info_id) (SELECT id FROM tbl_info)";
用於插入tbl_training的ID(有錯誤)
$sql4 = "INSERT INTO tbl_infotrainings (training_id) (SELECT id FROM tbl_trainings)";
所需的輸出在數據庫中將是:
+------+-----------+------------+
| id | info_id | training |
+------+-----------+------------+
| 1 | 1 | 1 |
+------+-----------+------------+
| 2 | 1 | 2 |
+------+-----------+------------+
| 3 | 1 | 3 |
+------+-----------+------------+
下面是完整的代碼:
if($_POST["Submit"]=="Submit"){
$sql1="INSERT INTO tbl_info VALUES ('NULL', '$fname', '$mname', '$lname', '$street', '$barangay', '$city', '$number', '$month', '$day', '$year', '$status' , '$spouse', '$dependent')";
$result1=mysql_query($sql1);
$sql3 = "INSERT INTO tbl_infotrainings (info_id) (SELECT id FROM tbl_info)";
$result3=mysql_query($sql3);
$sql4 = "INSERT INTO tbl_infotrainings (training_id) (SELECT id FROM tbl_trainings)";
$result4=mysql_query($sql4);
for ($i=0; $i<sizeof($checkbox);$i++){
$sql2="INSERT INTO tbl_trainings VALUES ('NULL', '".$checkbox[$i]."')";
$result2=mysql_query($sql2);
}
}
在此先感謝。
我認爲您錯過了一條有價值的信息來獲得一些幫助 - 您目前遇到的錯誤或不良結果是什麼? –
我真的不能告訴錯誤,因爲我只是在使用if else。但training_id不接受任何內容,僅在數據庫中爲0 – user1926313