PHP PDO多个查询没有任何价值 [英] PHP PDO Multiple queries doesnt get any value

问题描述

在此代码之前，我已经进行了内部联接，但是每次我有多个查询时，它似乎都不起作用.但是，当我执行单个查询时，它似乎正常工作.请高手帮帮我.谢谢！

I already did inner join before this code but then it seems like every time i have multiple queries it doesnt work. But then when I did it single query it seems fine and working. Please help me experts. Thank you!

<?php
include('db.php');
include('function.php');
if(isset($_POST["personal_info_id"]))
{
    $id1 = $_POST["personal_info_id"];
    $id2 = $_POST["personal_info_id"];
    $sql = "SELECT * FROM hr_details WHERE personal_info_id = '$id1'";
    $sql .="SELECT * FROM personal_info WHERE personal_info_id = '$id2'";
    $output = array();  
    $statement = $connection->prepare($sql);
    $statement->execute();
    $result = $statement->fetchAll();
    foreach($result as $row)
    {
        $output["firstname"] = $row["firstname"];
        $output["middlename"] = $row["middlename"];
        $output["lastname"] = $row["lastname"];
        $output["address"] = $row["address"];
        $output["birthdate"] = $row["birthdate"];
        $output["gender"] = $row["gender"];
        $output["religion"] = $row["religion"];
        $output["civil_status"] = $row["civil_status"];
        $output["biometrics"] = $row["biometrics"];
        if($row["image"] != '')
        {
        $output['user_image'] = '<img src="upload/'.$row["image"].'" class="img-thumbnail" width="50" height="35" /><input type="hidden" name="hidden_user_image" value="'.$row["image"].'" />';
        }
        else
        {
          $output['user_image'] = '<input type="hidden" name="hidden_user_image" value="" />';
         }

    }

    echo json_encode($output);

}
?>

推荐答案

尽管可以运行一次执行多个查询通过PDO进行呼叫，这样做的理由不单一.您的情况都不例外.它永远不会像您想的那样工作.不需要像这样在一个调用中填充多个查询，而是需要使用JOIN进行单个查询.

Although it's possible to run multiple queries in one call with PDO, there is not a single reason to do so. Neither your case is an exception. It will never work as you think. Instead of stuffing several queries in one call like this, you need a single query with JOIN.

您的代码的另一个问题是关于货物的备案声明.它看起来像一个真实的东西，却什么也没有保护.您应该在准备好的查询中使用参数.

Another issue with your code is a cargo cult prepared statement. It looks like a real one but protects nothing. You should use parameters in your prepared query.

$sql = "SELECT * FROM hr_details hd, personal_info pi WHERE 
        hd.personal_info_id=pi.personal_info_id AND hd.personal_info_id=?";
$statement = $connection->prepare($sql);
$statement->execute([$_POST["personal_info_id"]]);
$result = $statement->fetchAll();

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