Question

我已经阅读了几个SO线程，可以使用INNER JOIN更新查询更新两个数据库，但我无法让它工作，它只是抛出错误：

致命错误：未捕获的异常＆＃39; PDOException＆＃39;消息＆＃39; SQLSTATE [23000]：完整性约束违规：1052列＆＃39;类别＆＃39;在字段列表中是模糊的＆＃39;在C：\ xampp \ htdocs \ update.php：79堆栈跟踪：＃0 C：\ xampp \ htdocs \ update.php（79）：PDOStatement-＆gt;执行（数组）＃1 {main}抛出C：\第79行的xampp \ htdocs \ update.php

第79行： $result = $stmt->execute($prepare);

if (isset($_POST['update'])) {
    $category = isset($_POST['category']) ? $_POST['category'] : null;
    $manufactuer = isset($_POST['manufactuer']) ? $_POST['manufactuer'] : null;
    $model = isset($_POST['model']) ? $_POST['model'] : null;
    $serial = isset($_POST['serial']) ? $_POST['serial'] : null;
    $itemcondition = isset($_POST['itemcondition']) ? $_POST['itemcondition'] : null;
    $locationb = isset($_POST['locationb']) ? $_POST['locationb'] : null;
    $locationr = isset($_POST['locationr']) ? $_POST['locationr'] : null;
    $comments = isset($_POST['comments']) ? $_POST['comments'] : null;
    $purchased = isset($_POST['purchased']) ? $_POST['purchased'] : null;
    $retired = isset($_POST['retired']) ? $_POST['retired'] : null;
    $stolen = isset($_POST['stolen']) ? $_POST['stolen'] : null;

    $sql_part = array();
    $prepare = array();
    if ($category){
        $sql_part[] = 'category = :category';
        $prepare[':category'] = $category;
    }
    if($manufactuer){
        $sql_part[] = 'manufactuer = :manufactuer';
        $prepare[':manufactuer'] = $manufactuer;
    }
    if($model){
        $sql_part[] = 'model = :model';
        $prepare[':model'] = $model;
    }
    if($serial){
        $sql_part[] = 'serial = :serial';
        $prepare[':serial'] = $serial;
    }
    if($itemcondition){
        $sql_part[] = 'itemcondition = :itemcondition';
        $prepare[':itemcondition'] = $itemcondition;
    }
    if($locationb){
        $sql_part[] = 'locationb = :locationb';
        $prepare[':locationb'] = $locationb;
    }
    if($locationr){
        $sql_part[] = 'locationr = :locationr';
        $prepare[':locationr'] = $locationr;
    }
    if($comments){
        $sql_part[] = 'comments = :comments';
        $prepare[':comments'] = $comments;
    }
    if($purchased){
        $sql_part[] = 'purchased = :purchased';
        $prepare[':purchased'] = $purchased;
    }
    if($retired){
        $sql_part[] = 'retired = :retired';
        $prepare[':retired'] = $retired;
    }
    if($stolen){
        $sql_part[] = 'stolen = :stolen';
        $prepare[':stolen'] = $stolen;
    }
    $prepare[':barcode'] = $barcode;

    if(count($sql_part)){
        $sql = 'UPDATE assets a INNER JOIN assets_history b ON (a.barcode = b.barcode) SET ';
        $sql .= implode(', ', $sql_part);
        $sql .= ' WHERE a.barcode = :barcode AND b.barcode = :barcode';

        $stmt = $conn->prepare($sql);

        if($stmt){
            $result = $stmt->execute($prepare);
            $count = $stmt->rowCount();
            header('Location: ./usearch.php');
            exit;
        }
    }
}

这是数据库结构，如果需要：

  `barcode` int(6) UNSIGNED ZEROFILL NOT NULL
  `category` text NOT NULL
  `manufactuer` text NOT NULL
  `model` varchar(255) NOT NULL
  `serial` varchar(255) NOT NULL
  `itemcondition` text NOT NULL
  `locationb` text NOT NULL
  `locationr` text NOT NULL,
  `comments` varchar(255) NOT NULL
  `purchased` varchar(30) NOT NULL
  `retired` varchar(30) NOT NULL
  `stolen` varchar(30) NOT NULL

是否有更好的方法可以做到这一点，或者我错过了一些愚蠢的事情。

我也刚刚看到了在PHPMyAdmin中关联两个表格列的可能性，但尚未尝试过。这两个表是相同的，只会保留对记录的所有更新。

我对这些文章没有运气，其中一篇是我的代码的基础。

MySQL UPDATE syntax with multiple tables using WHERE clause

MySql update two tables at once

How to update two tables in one statement in SQL Server 2005?

MySQL, update multiple tables with one query

Answer 1

使用table name列category列表

if ($category){
    $sql_part[] = 'assets.category = :category';
    $prepare[':category'] = $category;
}

Answer 2

在我看来，资产＆＃39; ＆安培; ＆＃39; assets_history＆＃39;两者都有＆＃39;类别＆＃39;领域。因此，请更新您的代码:(用您的确切table_name替换＆＃39; assets＆＃39;或＆＃39; assets_history＆＃39;

if ($category){
        $sql_part[] = '<TABLE_NAME>.category = :category';
        $prepare[':category'] = $category;
    }

Answer 3

您必须附加表名或其别名，以避免像这样的混淆

if ($category){
        $sql_part[] = 'a.category = :category';
        $prepare[':category'] = $category;
    }
    if($manufactuer){
        $sql_part[] = 'a.manufactuer = :manufactuer';
        $prepare[':manufactuer'] = $manufactuer;
    }
    if($model){
        $sql_part[] = 'a.model = :model';
        $prepare[':model'] = $model;
    }
    if($serial){
        $sql_part[] = 'a.serial = :serial';
        $prepare[':serial'] = $serial;
    }
    if($itemcondition){
        $sql_part[] = 'a.itemcondition = :itemcondition';
        $prepare[':itemcondition'] = $itemcondition;
    }
    if($locationb){
        $sql_part[] = 'a.locationb = :locationb';
        $prepare[':locationb'] = $locationb;
    }
    if($locationr){
        $sql_part[] = 'a.locationr = :locationr';
        $prepare[':locationr'] = $locationr;
    }
    if($comments){
        $sql_part[] = 'a.comments = :comments';
        $prepare[':comments'] = $comments;
    }
    if($purchased){
        $sql_part[] = 'a.purchased = :purchased';
        $prepare[':purchased'] = $purchased;
    }
    if($retired){
        $sql_part[] = 'a.retired = :retired';
        $prepare[':retired'] = $retired;
    }
    if($stolen){
        $sql_part[] = 'a.stolen = :stolen';
        $prepare[':stolen'] = $stolen;
    }
    $prepare[':barcode'] = $barcode;

编辑：您也可以这样做来更新所有列

$columns = array('category','manufactuer','model','serial','itemcondition','locationb','locationr','comments','purchased','retired','stolen');
$sql_part = array();
foreach($columns as $column){
        $sql_part[] = 'a.'.$column.' = :'.$column;
        $sql_part[] = 'b.'.$column.' = :'.$column;
        $prepare[':'.$column] = '$'.$column;

}

这将为您节省几行代码

更新两个MySQL数据库

3 个答案: