Question

我只想插入数据库，但是下面的代码没有显示任何错误，也没有显示我在数据库中插入的数据。任何人都可以帮我调试这段代码。

<?php
$tips = $_POST["tips"];
$day = $_POST["day"];
$month = $_POST["month"];
$year = $_POST["year"];
$date = $year."-".$month."-".$day;

//echo "$date: $tips";

//========================
//  DataBase Connectivity
//========================
$con=mysqli_connect("localhost","root","","mydb");

// Check connection
if (mysqli_connect_errno($con))
  {
    echo "Failed to connect to Database";
    exit;
  }

else{
// Perform INSERT query
    mysqli_query($con,"INSERT INTO daily_tips (PublishingDate,Tips) VALUES ('$date','$tips')");
    echo "data inserted"
    exit;
}
mysqli_close($con);
?>

Answer 1

试

if( mysqli_query($con,"INSERT INTO daily_tips (PublishingDate,Tips) VALUES ('$date','$tips')"))
{
    echo 'success';
}else{
    echo 'failed';
}

Answer 2

确保您的日期格式类似于'2013-12-05'（$ date = $ year。“ - ”。$ month。“ - ”。$ day）

并尝试以下代码：

mysqli_query（$ con，“INSERT INTO daily_tips（PublishingDate，Tips）VALUES（'”。$ date。“'，'”。$ tips。“'）”）; echo“数据插入”; 出口;

Answer 3

试试这个

mysqli_query($con,"INSERT INTO daily_tips (PublishingDate,Tips) VALUES ('".$date."','".$tips."')");

Answer 4

首先你不需要提供else条件删除“else”块

如果一切正常，请尝试以下方法：

 $sql= mysqli_query($con,"INSERT INTO daily_tips (PublishingDate,Tips) VALUES  ('".$date."','".$tips."')");

 if( $sql) {
 echo "data inserted";
 } else {
 echo "not inserted";
 }

Answer 5

只需使用完整的代码并告诉我，它的工作正常，让我知道您为PublishingDate使用的数据类型是什么？

<?php
$tips = "Today tips";
$day = "12";
$month = "01";
$year = "2012";
$date = $year."-".$month."-".$day;

//echo "$date: $tips";

//========================
//  DataBase Connectivity
//========================
$con=mysqli_connect("localhost","root","","test");

// Check connection
if (mysqli_connect_errno($con))
  {
    echo "Failed to connect to Database";
    exit;
  }

else{
// Perform INSERT query
    mysqli_query($con,"INSERT INTO daily_tips (PublishingDate,Tips) VALUES ('$date','$tips')");
    echo "data inserted";
    exit;
}
mysqli_close($con);
?>

Answer 6

尝试此查询

mysqli_query($con,"INSERT INTO daily_tips (PublishingDate,Tips) VALUES (".$date.",".$tips.")");

MySQL INSERT查询有效，但未插入数据

6 个答案: