Question

我正在尝试使用表单更新数据库，但它没有添加项目或更新。删除项目有效，但就是这样。我做错了什么，我该怎么做才能解决这个问题？

enter image description here 非常感谢您的帮助。

INSERT ITEM PAGE

  <?php 
// Parse the form data and add inventory item to the system
if (isset($_POST['PART_DESC'])) {


             $PART_DESC = $row["PART_DESC"];
             $SERIAL_NUM = $row["SERIAL_NUM"];
             $RACK_NUM = $row["RACK_NUM"];
             $PART_TYPE_ID = $row["PART_TYPE_ID"];
             $PART_TYPE_DESC = $row["PART_TYPE_DESC"];
    // See if that product name is an identical match to another product in the system
    $sql = mysql_query("SELECT PART_ID FROM PART WHERE ='$PART_ID' LIMIT 1");
    $productMatch = mysql_num_rows($sql); // count the output amount
    if ($productMatch > 0) {
        echo 'Sorry you tried to place a duplicate "Product Name" into the system, <a href="inventory.php">click here</a>';
        exit();
    }
    // Add this product into the database now
    $sql = mysql_query("INSERT INTO PART (PART_DESC, SERIAL_NUM, RACK_NUM, PART_TYPE_DESC, LOCATION) 
        VALUES('$PART_DESC','$SERIAL_NUM','$RACK_NUM','$PART_TYPE_ID','$PART_TYPE_DESC',now() )") or die (mysql_error());
     $pid = mysql_insert_id();

    exit();
}

?>

<?php 
// Script Error Reporting
error_reporting(E_ALL);
ini_set('display_errors', '1');
?>

编辑项目页

<?php 
// Gather this product's full information for inserting automatically into the edit form below on page
if (isset($_GET['pid'])) {
    $targetID = $_GET['pid'];
    $sql = mysql_query("SELECT PART_ID, PART_DESC, SERIAL_NUM, RACK.RACK_NUM, PART.PART_TYPE_ID, PART_TYPE_DESC, LOCATION
FROM PART
INNER JOIN PART_TYPE ON PART.PART_TYPE_ID = PART_TYPE.PART_TYPE_ID
INNER JOIN RACK ON RACK.RACK_NUM = PART.RACK_NUM
WHERE PART_ID='$targetID' LIMIT 1");
    $productCount = mysql_num_rows($sql); // count the output amount
    if ($productCount > 0) {
        while($row = mysql_fetch_array($sql)){ 

             $id = $row["PART_ID"];
             $PART_DESC = $row["PART_DESC"];
             $SERIAL_NUM = $row["SERIAL_NUM"];
             $RACK_NUM = $row["RACK_NUM"];
             $PART_TYPE_ID = $row["PART_TYPE_ID"];
             $PART_TYPE_DESC = $row["PART_TYPE_DESC"];
             $LOCATION = $row["LOCATION"];
        }
    } else {
        echo "Sorry dude that crap dont exist.";
        exit();
    }
}
?>
<?php 
// Parse the form data and add inventory item to the system
if (isset($_POST['PART_DESC'])) {

    $pid = mysql_real_escape_string($_POST['thisID']);

             $PART_DESC = $row["PART_DESC"];
             $SERIAL_NUM = $row["SERIAL_NUM"];
             $RACK_NUM = $row["RACK_NUM"];
             $PART_TYPE_ID = $row["PART_TYPE_ID"];
             $PART_TYPE_DESC = $row["PART_TYPE_DESC"];
             $LOCATION = $row["LOCATION"];

    // See if that product name is an identical match to another product in the system
    $sql = mysql_query("UPDATE PART SET PART_DESC='$PART_DESC', SERIAL_NUM='$SERIAL_NUM', PART.PART_TYPE_DESC='$PART_TYPE_DESC', RACK.RACK_NUM='$RACK_NUM', LOCATION='$LOCATION' 
    INNER JOIN PART_TYPE ON PART.PART_TYPE_ID = PART_TYPE.PART_TYPE_ID
INNER JOIN RACK ON RACK.RACK_NUM = PART.RACK_NUM WHERE PART.PART_ID='$pid'");
    header("location: inventory.php"); 
    exit();
}

＆GT;

Answer 1

这段代码看起来很奇怪：

INSERT INTO PART 
  (PART_DESC, SERIAL_NUM, RACK_NUM, PART_TYPE_DESC, LOCATION) 
VALUES
('$PART_DESC','$SERIAL_NUM','$RACK_NUM','$PART_TYPE_ID','$PART_TYPE_DESC',now())"

parts表只有字段：

 part_id
 , part_desc
 , serial_num
 , rack_num
 , part_type_id

所以这些是你可以插入的唯一字段。外键不会神奇地将插页分散到您必须自己完成的多个表中外键仅会阻止您在表格中插入悬空数据（不链接到任何内容的FK），即IT。

如何通过一次插入仍然可以做到你想要的东西
如果您一次完成插入操作，（我推荐）执行以下操作：
创建一个blackhole表，其中包含您要插入的所有字段在黑洞表上创建一个after insert trigger并在其中执行插入逻辑这将自动将所有插入逻辑放在一个事务中，因此即使您使用的是MyISAM表，您仍然可以进行ACID安全交易： - ）。

/*create the blackhole table, adjust field types to match your tables*/
CREATE TABLE  `test`.`bhpartsinsert` (
  `part_desc` varchar(255) NOT NULL,
  `serial_num` varchar(45) NOT NULL,
  `part_type_desc` varchar(45) NOT NULL,
  `location` varchar(45) NOT NULL,
  PRIMARY KEY (`part_type_desc`)
) ENGINE=BLACKHOLE DEFAULT CHARSET=latin1;

DELIMITER $$

CREATE TRIGGER ai_bhparts_insert_each AFTER INSERT 
  ON bhpartsinsert FOR EACH ROW
BEGIN
  DECLARE mypart_type_id INTEGER;
  DECLARE myrack_num INTEGER;

  SELECT part_type_id INTO mypart_type_id FROM part_type 
    WHERE part_type_desc = new.part_type_desc LIMIT 1;
  IF mypart_type_id IS NULL THEN BEGIN 
    INSERT INTO part_type VALUES (null, new.part_type_desc);
    SELECT LAST_INSERT_ID() INTO mypart_type_id;
  END; END IF;

  SELECT rack_num INTO myrack_num FROM rack 
    WHERE location = new.location LIMIT 1;
  IF myrack_num IS NULL THEN BEGIN
    INSERT INTO rack VALUES (null, new.location);
    SELECT LAST_INSERT_ID() INTO myrack_num;
  END; END IF;

  INSERT INTO part VALUES 
    (null, new.part_desc, new.serial_num, myrack_num, mypart_type_id);
END $$

DELIMITER ;

如何运作
无论你插入黑洞表中的是什么都会被遗忘，但触发器会被解雇所以我做了一个触发器，每次插入每一行后都会触发触发器确保机架中有一个位置，并确保有一个part_type，然后在零件表中正确插入。
我假设您的主键字段是自动递增整数。如果您插入null MySQL将为您创建一个独特的PK 仅使用null，0有时可行，有时不依赖于您的配置，null始终有效

BTW，请不要使用全部大写字母作为你的字段名称，如果你已经习惯了SQL语法，其中KEYWORDS都是大写的，这会让你头疼吗

PHP / MYSQL如何插入/更新多个表

1 个答案: