Question

我使用XAMPP在mysql中创建了一个名为test226333的数据库我正在尝试将数据发送到这样的表中 “http://localhost/write_data.php?value=100”
for database and table details screenshot click here

我的PHP代码是

<?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);


?>

进入链接时没有显示任何内容，数据也没有存储在数据库中

请帮助我提前谢谢..

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();

使用XAMPP数据不会存储在mySQL数据库中

1 个答案: