Question

我有一个生成并运行mysql更新查询的函数。当我在phpMyAdmin中运行生成的代码时，我没有麻烦，但是当我尝试从我的网站运行它时，我收到一个错误。奇怪的是我使用相同的格式进行其他几个更新查询，并且它们正在完美地运行。

MySQL查询：

UPDATE `capc`.`bio_positions` SET `Pos_Name` = 'IT Specialist' WHERE `bio_positions`.`Pos_ID` = 63;UPDATE `capc`.`bio_positions` SET `Pos_Company` = 'CSG' WHERE `bio_positions`.`Pos_ID` = 63;

错误：

Could not get data: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'UPDATE `capc`.`bio_positions` SET `Pos_Company` = 'CSG' WHERE `bio_position' at line 1

用于生成MySQL查询的代码

function update($rank, $name, $company) {
    if (!empty($rank) && $rank !== $this->Pos_Rank) {
        $sql = "UPDATE `capc`.`bio_positions` SET `Pos_Rank` = " . $rank . " WHERE `bio_positions`.`Pos_ID` = " . $this->Pos_ID . ";";
    }
    if (!empty($name) && $name !== $this->Pos_Name) {
        $sql .= "UPDATE `capc`.`bio_positions` SET `Pos_Name` = '" . $name . "' WHERE `bio_positions`.`Pos_ID` = " . $this->Pos_ID . ";";
    }
    if (!empty($company) && $company !== $this->Pos_Company) {
        $sql .= "UPDATE `capc`.`bio_positions` SET `Pos_Company` = '" . $company . "' WHERE `bio_positions`.`Pos_ID` = " . $this->Pos_ID . ";";
    }
    echo "<br>" . $sql . "<br>";
    if (!empty($sql)) {
        $capc = new CAPC;
        $capc->query($sql);
        Bio_Position($this->Pos_ID);
    }
}

从有效的答案更新更新功能

function update($rank, $name, $company) {
    $capc = new CAPC;
    $sql = "UPDATE `capc`.`bio_positions` SET ";
    $go = 0;
    if (!empty($rank) && $rank !== $this->Pos_Rank) {
        $sql .= " `Pos_Rank` = " . $rank;
        $go++;
    }
    if (!empty($name) && $name !== $this->Pos_Name) {
        if($go > 0){
            $comma = ",";
        }
        $sql .= $comma . " `Pos_Name` = '" . $name . "'";
        $go++;
    }
    if (!empty($company) && $company !== $this->Pos_Company) {
        if($go > 0){
            $comma = ", ";
        }
        $sql .= $comma . " `Pos_Company` = '" . $company . "'";
        $go++;
    }
    $sql .= " WHERE `bio_positions`.`Pos_ID` = " . $this->Pos_ID . ";";

    if (!empty($sql) && $go > 0) {
        //echo $sql . "<br>";
        $capc = new CAPC;
        $capc->query($sql);
    }
}

Answer 1

当我在phpMyAdmin中运行生成的代码时，我没有麻烦

更新capc。bio_positions SET Pos_Name ='IT专家'在哪里   bio_positions。Pos_ID = 63;

更新capc。bio_positions设置Pos_Company ='CSG'WHER bio_positions。Pos_ID = 63;

虽然phpMyAdmin似乎一次执行多个查询（它实际上是在后端为你分割它们），但在你的代码中你需要单独执行它们或重写SQL来更新单个列中的多个列声明，即UPDATE capc.bio_positions SET Pos_Name = 'IT Specialist', Pos_Company = 'CSG' WHERE bio_positions.Pos_ID = 63。

如果你正在使用mysqli，你也可以查看mysqli::multi_query。

Answer 2

至少有一些MySQL API不允许在一个“批处理”中运行多个查询，但PHPMyAdmin允许它。

看到查询更新了同一组行，您可以将其重写为单个查询，这样可以加快执行速度，因为WHERE条件只需要评估一次;

UPDATE `capc`.`bio_positions` 
SET `Pos_Name` = 'IT Specialist', `Pos_Company` = 'CSG' 
WHERE `bio_positions`.`Pos_ID` = 63

Answer 3

您的WHERE具有相同的条件，那么为什么要将此分为三个查询？你可以这样做：

UPDATE `capc`.`bio_positions` SET `Pos_Rank` = " . $rank . ", `Pos_Name=` = " . $name . ", Pos_Company=" . $company . " WHERE `bio_positions`.`Pos_ID` = " . $this->Pos_ID;

编辑:(根据评论）

function update($rank, $name, $company) {
       $new_rank = $this->Pos_Rank;
       $new_name = $this->Pos_Name;
       $new_company = $this->Pos_Company;

        if (!empty($rank)) {
            $new_rank = $rank;
        }
        if (!empty($name)) {
                $new_name = $name;
        }
        if (!empty($company)) {
                $new_company = $company;
        }

        $q = ...

}

Answer 4

Try this    

UPDATE `capc`.`bio_positions` 
    SET `Pos_Name` = 'IT Specialist' 
    WHERE `Pos_ID` = 63;

在站点中运行时出现SQL错误

4 个答案: