使用XAMPP

Question

我一直在使用XAMPP命名test226333我试图将数据发送到表像这样
“http://localhost/write_data.php?value=100”
for database and table details screenshot click here

我的PHP代码中创建在MySQL数据库中的数据是没有得到存储在MySQL数据库，同时进入链接和数据也没有得到存储在数据库中

<?php 

$dbusername = "ganesh"; 
$dbpassword = "varma"; 
$server = "localhost"; 

$dbconnect = mysql_connect($server, $dbusername, $dbpassword); 
$dbselect = mysql_select_db("test226333",$dbconnect); 



$sql = "INSERT INTO test226333.data VALUES ('".$_GET["value"]."')";  


mysql_query($sql); 


?> 

没有什么是显示

请帮我在这提前
谢谢..

Answer 1

试试这个：

$servername = "localhost"; 
    $username = "ganesh"; 
    $password = "varma"; 
    $dbname = "test226333"; 
    $mysqli = new mysqli($servername, $username, $password, $dbname); 
    if ($mysqli->connect_errno) { 
     echo "Failed to connect to MySQL: (" . $mysqli->connect_errno . ") " . $mysqli->connect_error; 
    } 

    /* Prepared statement, stage 1: prepare */ 
    if (!($stmt = $mysqli->prepare("INSERT INTO <your table name>(col1, col2, col3) VALUES (?, ?, ?)"))) { 
     echo "Prepare failed: (" . $mysqli->errno . ") " . $mysqli->error; 
    } 

    /* Prepared statement, stage 2: bind and execute */ 

    if (!$stmt->bind_param("i", $val1)) { 
     echo "Binding parameters failed: (" . $stmt->errno . ") " . $stmt->error; 
    } 
if (!$stmt->bind_param("s", $val2)) { 
     echo "Binding parameters failed: (" . $stmt->errno . ") " . $stmt->error; 
    } 
if (!$stmt->bind_param("i", $val3)) { 
     echo "Binding parameters failed: (" . $stmt->errno . ") " . $stmt->error; 
    } 

    if (!$stmt->execute()) { 
     echo "Execute failed: (" . $stmt->errno . ") " . $stmt->error; 
    } 

    /* explicit close recommended */ 
    $stmt->close();