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Php 在两个表中插入值_Php_Mysql_Sql - Fatal编程技术网

Php 在两个表中插入值

Php 在两个表中插入值,php,mysql,sql,Php,Mysql,Sql,完成表单后,我创建了一个PDF,并将表单的一些值存储到一个表中。我想将表单值存储到一个单独的表中。我无法将代码存储到两个表中。这是我的工作代码,没有添加第二个SQL查询: <?php session_start(); $errors = array(); include('connect.php'); //if($_POST['email'] == ''){ // $errors[]= 'Email cannot be empty'; //} if( count($e

完成表单后,我创建了一个PDF,并将表单的一些值存储到一个表中。我想将表单值存储到一个单独的表中。我无法将代码存储到两个表中。这是我的工作代码,没有添加第二个SQL查询:

<?php
session_start(); 
$errors = array(); 
include('connect.php'); 

//if($_POST['email'] == ''){
//    $errors[]= 'Email cannot be empty';  
//}


if( count($errors) === 0 ) 
{
    $description = pathinfo(__FILE__, PATHINFO_FILENAME);  
    $filename = date('Ymdhis').".pdf"; 
    $username = $_SESSION['username']; 
    $sql = "INSERT INTO documents (id, userid, description, name, date) VALUES (NULL, {$_SESSION['id']}, '{$description}' ,'{$filename}', NOW())"; 
     $success = mysql_query($sql);    

    include('../MPDF/mpdf.php'); 
    ob_start(); 
    include('../documents/siteSubmit.php');
    $html = ob_get_contents(); 
    ob_end_clean(); 
    ob_end_flush(); 

    $mpdf=new mPDF();
    $folder = "../userDocuments/{$_SESSION['username']}"; 
    if(!is_dir($folder))
    {
        mkdir($folder, 0755, true);  
    } 
    $mpdf->writeHTML($html); 
    $mpdf->Output('../userDocuments/'.$_SESSION['username'].'/'.$filename); 
    echo json_encode(array('succes' => true)); 
}
else 
{
    echo json_encode(array('errors' => $errors)); 
}

您可以仅使用此查询,但使用不同的名称。那样

  $sql = "INSERT INTO documents (id, userid, description, name, date) VALUES (NULL, {$_SESSION['id']}, '{$description}' ,'{$filename}', NOW())"; 
  $success = mysql_query($sql);  
  $sql2 = "INSERT INTO sitesubmit (first_name, last_name, email, telephone, truck_number, truck_mileage, carrier, site_number, lat, longitude, comments, job_completion) VALUES ('first_name', 'last_name', 'email', 'telephone', 'truck_number', 'truck_mileage', 'carrier', 'site_number', 'lat', 'longitude ', 'comments','job_completion')"; 

 $success2 = mysql_query($sql2);  
编辑:

第二个查询缺少
作业完成
的值,因此此查询将失败


您必须对该列进行赋值。

这不起作用。当我这样做时,它只会添加到第一个表中,而忽略第二个表。@NichoDiaz它会起作用的,你试过了吗?请注意,我已经更改了查询的名称。是的,我尝试了它,并用您的代码再次尝试,以确保我没有弄糟。还是不行。我不知道我在这里遗漏了什么。请重试我的代码,我编辑并添加了一个值,用于
job\u completion
好的,谢谢!它输入了数据库!伟大的唯一的问题是它输入了我们在数据库中键入的实际值,而不是将输入的值抓取到表单的输入字段中。。。。每个值都与表单上的name='telephone'相同。。。我做错了什么?
  $sql = "INSERT INTO documents (id, userid, description, name, date) VALUES (NULL, {$_SESSION['id']}, '{$description}' ,'{$filename}', NOW())"; 
  $success = mysql_query($sql);  
  $sql2 = "INSERT INTO sitesubmit (first_name, last_name, email, telephone, truck_number, truck_mileage, carrier, site_number, lat, longitude, comments, job_completion) VALUES ('first_name', 'last_name', 'email', 'telephone', 'truck_number', 'truck_mileage', 'carrier', 'site_number', 'lat', 'longitude ', 'comments','job_completion')"; 

 $success2 = mysql_query($sql2);