Question

使用MySql和phpmyadmin，我有两个表person和report，它们使用字段person_id作为report表中的外键进行链接。使用一个PHP表单我试图将人员详细信息添加到person表。使用相同的表单，我希望它将日期发布到report表，然后生成一个自动增量report_id，并使用person_id外国人将该人与该日期的特定报告相关联键。

这是我目前的代码

<?php

if (isset($_POST['submitted'])) {

    include('dbcon.php'); //link to connection file

    $pid = "SELECT person_id FROM person\n"
         . "ORDER BY person_id DESC\n"
         . "LIMIT 1"; //variable finds last generated person_id

    $sqlinsert1 = "INSERT INTO person (person.title, person.first_name, person.last_name,  person.address, person.contact_no, person.email, person.ha_id) VALUES ('$_POST[title]' , '$_POST[first_name]' , '$_POST[last_name]' , '$_POST[address]' , '$_POST[contact_no]' , '$_POST[email]' , '$_POST[ha_id]')";

    $sqlinsert2 = "INSERT INTO report (report.date) VALUES ('$_POST[date]') WHERE ($pid = report.person_id)";

    if (!mysqli_query($dbcon, $sqlinsert1)) {
        die('Error inserting record');
    } //end of nested if statement

    if (!mysqli_query($dbcon, $sqlinsert2)) {
        die('Error inserting record');
    } //end of nested if statement

    $newrecord = "new record added to database";
} //end of if statement

?>

我已经创建了变量$pid，它将找到person_id表中生成的最后一个person，我已对此进行了测试，并且它在phpmyadmin中有效。我想使用此变量将日期与person_id相关联，并将其放入report表。

这可能听起来很复杂，但我确信有一个简单的答案。

Answer 1

我认为错误在$ _POST [date]中使用'引号'。你的代码如下......

当您插入一个人后，取用人名。

然后插入报告

$sqlinsert2 = "INSERT INTO report (report.pid,report.date) VALUES ('$pid',"$_POST['date']");//changed check it

然后你的两个表将用pid（引用键）链接在一起。

Answer 2

在与我的讲师交谈并花费数小时回应所有结果后，我们发现连接sqli语句（'mysqli_query'）连接到服务器但实际上并没有连接到数据库本身，因此不允许person_id使用select语句从person表中提取。

我们输入标准的sql连接语句并分别指定数据库名称。这似乎工作，我已经从另一个表（'defect_id'）添加了另一个外键，并为缺陷表添加了另一个插入语句。我现在可以运行此代码，它工作正常。

<?php

            if (isset($_POST['submitted'])) {

    include('dbcon.php');

            //insert statement 1    
        $sqlinsert1 = "INSERT INTO person (title, first_name, last_name, address, contact_no, email, ha_id) VALUES ('$_POST[title]' , '$_POST[first_name]' , '$_POST[last_name]' , '$_POST[address]' , '$_POST[contact_no]' , '$_POST[email]' , '$_POST[ha_id]')";

        if (!mysqli_query($dbcon, $sqlinsert1)) {
        die('Error inserting record1');
        } //end of nested if statement1


                    // connect to database      
$dbcon2 = mysql_connect('localhost' , 'root' , '' ); //mysqli query would not connect to db so using mysql connection

        if (!$dbcon2){
        die('error connecting to database');
        }

$dbselect = @mysql_select_db('potholes_v2');

        if (!$dbselect){
        die('error connecting database');
        }

                    //insert statement 2            
$sqlinsert2 = "INSERT INTO defect (road, location, carriageway, lane, diameter, depth, speed, description) VALUES ('$_POST[road]' , '$_POST[location]' , '$_POST[carriageway]' , '$_POST[lane]' , '$_POST[diameter]' , '$_POST[depth]' , '$_POST[speed]' , '$_POST[description]')";

        if (!mysqli_query($dbcon, $sqlinsert2)) {
        die('Error inserting record2');
        } //end of nested if statement1


mysql_close(); //close database connection


                   //connect to database
$dbcon2 = mysql_connect('localhost' , 'root' , '' ); //mysqli query would not connect to db so using mysql connection

        if (!$dbcon2){
        die('error connecting to host');
        }

$dbselect = @mysql_select_db('potholes_v2');

        if (!$dbselect){
        die('error connecting database');
        }

                    //select person_id value
$value = mysql_query("SELECT person_id FROM person ORDER BY person_id DESC" , $dbcon2); //selects last entry of person_id in person table

        if (!$value) {
        die ('error, no values'); //error if no value selected
        }

$value2 = mysql_result($value,0); //specifies new value to be inserted into report table as person_id foreign key

                    //select defect_id value    
$defect = mysql_query("SELECT defect_id FROM defect ORDER BY defect_id DESC" , $dbcon2); //selects last defect_id entry from defect table

        if (!$defect) {
        die ('Error, no values for defect'); //error if no value selected
        }

$defect2 = mysql_result($defect,0); //specifies new defect value to be placed in report table

                    //insert statement 3
$sqlinsert3 = "INSERT INTO report (date, person_id, defect_id) VALUES ('$_POST[date]' , $value2 , $defect2)"; //inserts date, person_id and defect_id values into report table

        if (!mysqli_query($dbcon, $sqlinsert3)) {
        die('Error inserting record 3');
        } //end of nested if statement1


mysql_close(); //close database connection  





$newrecord = "new record added to database";  //gives feedback for successful submission

}  //end of initial if statement

?>

使用单个php表单将数据添加到两个表

2 个答案: