Question

我理解有一种方法可以从select语句中插入一个常量，我从here找到了来源，如：

INSERT INTO MyTable(ColA,ColB,ColC)
SELECT 1,colBB,colCC FROM MyTable2

但是是否可以添加用户输入值（使用php）而不是常量值？如果可能，请提供示例。提前谢谢。

更新：我试图创建一个简单的网页，但有一些语法错误，我不知道解决它：

解析错误：语法错误，意外''（T_ENCAPSED_AND_WHITESPACE），期望标识符（T_STRING）或变量（T_VARIABLE）或数字（T_NUM_STRING）在第10行的C：\ xampp \ htdocs \ webservice \ result.php

以下是我的代码：

<?php
//start a session

    require("config.inc.php");
    $username = $_SESSION["username"];
    if(!empty($_POST)){
    //check if user choose non-required drop down list
        if(empty($_POST['subcategory'])){
            if(empty($_POST['yearofstudy'])) {
                $query = "INSERT INTO comments ( username, title, message, tousername ) 
                          SELECT :username, :title, :message, username 
                          FROM utarstudents WHERE faculty = :faculty";
                          $query_params = array(
                            ':faculty' => $_POST['category'],
                            ':username' =>$username,
                            ':title' =>$_POST['title'],
                            'message' =>$_POST['message']

                            );
            }
            else {
                $query = "INSERT INTO comments ( username, title, message, tousername ) 
                          SELECT :username, :title, :message, username 
                          FROM utarstudents WHERE faculty = :faculty AND year_of_study = :yearofstudy";
                          $query_params = array(
                            ':faculty' => $_POST['category'],
                            'yearofstudy' => $_POST['yearofstudy'],
                            ':username' =>$username,
                            ':title' =>$_POST['title'],
                            'message' =>$_POST['message']
                            );
            }
        }
        else {
            if(empty($_POST['yearofstudy'])) {
                $query = "INSERT INTO comments ( username, title, message, tousername ) 
                           SELECT :username, :title, :message, username 
                          FROM utarstudents WHERE faculty = :faculty AND course = :course";
                          $query_params = array(
                            ':faculty' => $_POST['category'],
                            ':course' => $_POST['subcategory'],
                            ':username' =>$username,
                            ':title' =>$_POST['title'],
                            'message' =>$_POST['message']
                            );
            }
            else {
                $query = "INSERT INTO comments ( username, title, message, tousername ) 
                           SELECT :username, :title, :message, username 
                          FROM utarstudents WHERE faculty = :faculty AND year_of_study = :yearofstudy AND course = :course";
                          $query_params = array(
                            ':faculty' => $_POST['category'],
                            'yearofstudy' => $_POST['yearofstudy'],
                            ':course' => $_POST['subcetagory'],
                            ':username' =>$username,
                            ':title' =>$_POST['title'],
                            'message' =>$_POST['message']
                            );

            }

        }

         try {
            $stmt   = $db->prepare($query);
            $result = $stmt->execute($query_params);
        }
        catch (PDOException $ex) {
            die("Failed to run query: " . $ex->getMessage());
        }
    }

Answer 1

以下是我编辑过的代码片段，可以完成您的尝试。所以在我的表单文件中，我有这样的东西来为项目做贡献：

<form name="contribute" method="post" action="contribute-dbquery.php" onsubmit="return validateForm()">
    First Name:
    <input name="nameValue" type="text" size="40" maxlength="12" required/>

    <input name="Submit" type="submit" value="Add"/>
</form>

所以它说的是当Add使用nameValue的值转到Tcontribute-dbquery.php时。然后在我的contrib-dbquery.php中，我将表单中的nameValue分配给$ name，然后将其插入到我的数据库中。我将它分配给变量，因为我也在该页面上使用过它。如果需要，您可以将其插入数据库中。

$name = $_POST['nameValue'];
$insert_sql = "INSERT INTO mastertable (name) VALUES (' " . $name . " ')";

如果这有助于将其标记为已回答。如果您需要任何帮助，请告诉我。

用select插入SQL数据库

1 个答案: