如何使用php从数据库中查询变量
大家好 您只是不执行查询。使用MySQLi:如何使用php从数据库中查询变量,php,html,database,Php,Html,Database,大家好 您只是不执行查询。使用MySQLi: <?php $servername = "localhost"; $username = "root"; $password = ""; $db = "dbthesis"; $conn = new mysqli($servername, $username, $password, $db); if ($conn->connect_error) { die("Connection failed: " . $conn-&g
<?php
$servername = "localhost";
$username = "root";
$password = "";
$db = "dbthesis";
$conn = new mysqli($servername, $username, $password, $db);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
if(isset($_POST['save'])){
$FName = $_POST['FName'];
$MName = $_POST['MName'];
$sql = "INSERT INTO tbstudinfo (Transaction_Number, First_Name, Middle_Name) VALUES ('000', '$FName', '$MName')";
if ($conn->query($sql) === TRUE) {
echo "Successfully Added";
} else {
echo "<p>Insertion Failed.</p>";
}
}
$conn->close();
您只进行查询,不运行查询。这是这个代码
$FName = $_POST['FName'];
$MName = $_POST['MName'];
$sql = "INSERT INTO tbstudioinfo (Transaction_Number, First_Name, Middle_Name) VALUES ('000','$FName','$MName')";
// code below runs your query
if (mysqli_query($conn, $sql)) {
echo "Successfully Added";
} else {
echo "Error: " . $sql . "<br>" . mysqli_error($conn);
}
$FName=$\u POST['FName'];
$MName=$_POST['MName'];
$sql=“插入tbstudioinfo(交易编号、名字、中间名)值('000'、'$FName'、'$MName')”;
//下面的代码运行您的查询
if(mysqli_查询($conn,$sql)){
echo“已成功添加”;
}否则{
echo“Error:”.$sql.
“.mysqli_Error($conn);
}
正如@executable所提到的,您正在代码中定义查询,但没有执行它
定义连接对象(Mysqli、PDO..)
准备查询并绑定变量
执行您的查询
下面是一个使用预置语句的示例
<?php
$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "myDB";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
if( isset($_POST['save']) ){
// prepare and bind
$stmt = $conn->prepare("INSERT INTO 'tbstudinfo' (Transaction_Number, First_Name, Middle_Name) VALUES (?, ?, ?)");
$stmt->bind_param("sss", $transaction_number, $FName, $MName);
// set parameters and execute
$transaction_number = '000';
$FName= $_POST['FName'];
$MName= $_POST['MName'];
$stmt->execute();
echo "Successfully Added";
}else{
echo "<p>Nothing Posted</p>";
}
代码已经发布了吗?先生,很抱歉我忘记添加代码了。请以文本形式发布您的代码。你需要执行你的查询这是我的php文件,先生。编辑您的问题以添加额外信息。不要将其放在注释中。不要忘记查询中$FName
和$MName
周围的引号。