无法使用PHP创建数据库,并且我的语法正确 [英] can not create a database with PHP and my syntax is correct

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问题描述

怎么不能在phpMyAdmin上创建数据库?我连接到PHP就很好了,没有错误消息.但是,当我尝试创建一个数据库(例如phpMyAdmin)时,我总是收到错误消息.如果有问题,我将在页面上放置一个<form>并提交到我的数据库代码所在的second.php文件中(以下显示second.php).

我几乎试图在phpMyAdmin中显示数据库,但是什么都没有显示:

<html>
    <body>
        <?php
            $con = mysqli_connect("localhost", "user", "password");
            if(!$con){
                echo "could not connect";
            } else {
                echo "good connection";
            }

            //creation of database
            $sql= 'CREATE DATABASE project';
            if(mysql_query($sql,$con)){
                echo 'DB created succesfully';
            } else {
                echo 'error creating DB' . mysql_errno();
            }
        ?>
    </body>
</html>

解决方案

一个问题是您与mysql i连接并且在其余时间都在使用mysql.选择1个库(不是mysql),并仅使用它.

if(mysqli_query($con, $sql)){

echo 'error creating DB' . mysqli_error($con);

How is it is can not create a database on phpMyAdmin? I connect to PHP just fine, no error message. But when I am trying to create a database so phpMyAdmin, I keep getting error. In case it matters I have one page with a <form> and has a submit to my second.php file where my database code is (second.php is shown below).

I am pretty much trying to display the database in phpMyAdmin but nothing shows up:

<html>
    <body>
        <?php
            $con = mysqli_connect("localhost", "user", "password");
            if(!$con){
                echo "could not connect";
            } else {
                echo "good connection";
            }

            //creation of database
            $sql= 'CREATE DATABASE project';
            if(mysql_query($sql,$con)){
                echo 'DB created succesfully';
            } else {
                echo 'error creating DB' . mysql_errno();
            }
        ?>
    </body>
</html>

解决方案

One issue is you connected with mysqli and are using mysql the rest of the time. Pick 1 (not mysql) library and use only that.

if(mysqli_query($con, $sql)){

and

echo 'error creating DB' . mysqli_error($con);

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