Question

通常我看到的原因是混合了mysql和mysqli或者列和字段不匹配的人，但这不是这里的情况。按下提交按钮时没有错误。错误报告显示没有错误。我已经尝试了一个提议的解决方案，它说要改变数据格式，但没有做任何修复它。

http://imgur.com/vh7OAvD＆amp; http://imgur.com/vTtEnSN应该澄清一下

指数：

<?php 
session_start();

?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN"
        "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en" lang="en">
<head>
    <meta http-equiv="content-type" content="text/html; charset=utf-8" />
    <title>Custom Version</title>
</head>
<body>
<form action="add.php" method="post">
    <p>
        <label for="project_number">Project Number:</label>
        <input type="number" name="project_number" id="project_number">
    </p>
    <p>
        <label for="project_name">Project Name:</label>
        <input type="text" name="project_name" id="project_name">
    </p>
    <p>
        <label for="project_desc">Project Description:</label>
        <input type="text" name="project_desc" id="project_desc">
    </p>
    <p>
        <label for="project_start">Project Start Date:</label>
        <input type="date" name="project_start" id="project_start">
    </p>
    <p>
        <label for="project_end">Project End Date:</label>
        <input type="date" name="project_end" id="project_end">
    </p>
    <input type="submit" value="Submit">
</form>
</body>
</html>

Add.php

<?php
session_start();
include_once("header.php");
?>
<html>
<body>
<?php 
mysqli_select_db($con,$DATABASE);

$sql="INSERT INTO project (project_number,project_name,project_desc,project_start,project_end)VALUES('$_POST[project_number]','$_POST[project_name]','$_POST[project_desc]','$_POST[project_start]','$_POST[project_end]')";

echo $sql;
if($con = new mysqli($SERVER, $USER, $PASS, $DATABASE)){ echo "success"; }else{ echo "error: " . mysqli_error($con); }
mysqli_close($con);
?>
</body>
</html>

的header.php

<?php
$SERVER = 'localhost';
$USER = 'user';
$PASS = 'pass';
$DATABASE = 'database';

error_reporting(E_ALL); ini_set('display_errors', 1);   
// Create connection
$con = new mysqli($SERVER, $USER, $PASS);

// Check connection
if ($con->connect_error) {
    die("Connection failed: " . $con->connect_error);
} 
echo "Connected successfully";
?>

更新：解决方案是通过添加：

来执行SQL

if ($con->query($sql) === TRUE) {
    echo "New record created successfully";
} else {
    echo "Error: " . $sql . "<br>" . $conn->error;
}

Answer 1

问题在于这条线，

$sql="INSERT INTO project (project_number,project_name,project_desc,project_start,project_end)
VALUES('$_POST'[project_number]','$_POST[project_name],'$_POST[project_desc],'$_POST[project_start]','$_POST[project_end]')";
              ^ see the single quote here             ^ missing quote here  ^ missing quote here

所以你的查询应该是这样的：

$sql="INSERT INTO project (project_number,project_name,project_desc,project_start,project_end)VALUES('$_POST[project_number]','$_POST[project_name]','$_POST[project_desc]','$_POST[project_start]','$_POST[project_end]')";

旁注：您的查询容易受到SQL注入的影响。始终准备，绑定并执行您的查询以防止任何类型的SQL注入。 And this is how you can prevent SQL injection in PHP

<强>更新

我如何修复日期格式？

这里是将日期格式更改为MySQL DATE类型的功能。

// 03/22/2000 ==> 2000-03-22
function mysql_date_format($date){
    $unixdatetime = DateTime::createFromFormat('m/d/Y', $date)->getTimestamp();
    $formatted_date = strftime("%Y-%m-%d",$unixdatetime);
    return $formatted_date;
}

现在使用 project_start 和 project_end 日期调用此函数，如下所示：

$project_start = mysql_date_format($_POST['project_start']);
$project_end = mysql_date_format($_POST['project_end']);

并在INSERT查询中使用这些格式化的值，如下所示：

$sql="INSERT INTO project (project_number,project_name,project_desc,project_start,project_end)VALUES('$_POST[project_number]','$_POST[project_name]','$_POST[project_desc]','$project_start','$project_end')";

Answer 2

试试这个，你错过了$_POST value $_POST['project_number']中的mysql日期格式和单引号，而不是$_POST[project_number]

<?php
session_start();
include_once("header.php");
?>
<html>
<body>
 <?php 
 $start_date = date('Y-m-d',strtotime($_POST['project_start']));
  $end_date = date('Y-m-d',strtotime($_POST['project_end']));
 mysqli_select_db($con,$DATABASE);


     $sql="INSERT INTO project (project_number,project_name,project_desc,project_start,project_end)VALUES('$_POST['project_number']','$_POST['project_name']','$_POST['project_desc']','$start_date','$end_date')";

mysqli_query($con,$sql);
mysqli_close($con);
 ?>

我错过了什么。 SQL插入不起作用。错误报告没有显示

2 个答案: