[英]My php code for connecting android application to phpMyAdmin database is not working
[英]Database and PHP not connecting to phpMyAdmin (basic code)
index.html的:
<!DOCTYPE html>
<html lang="en">
<head>
<body>
<form id="myform" action="userinfo.php" method="post" >
Name: <input type="test" name="name" autofocus>
Age: <input type="text" name="age" >
<button id="sub"> save</button>
</form>
<span id="result"></span>
<script src="script/jquery-1.8.1.min.js" type="text/javascript"></script>
<script src="script/my_script.js" type="text/javascript"></script>
</body>
</html>
my_script.js:
$("#sub").click( function() {
$.post( $("#myform").attr("action"),
$("#myform :input").serializeArray(),
function(info){ $("#result").html(info);
});
clearInput();
});
$("#myform").submit( function() {
return false;
});
function clearInput() {
$("#myform :input").each( function() {
$(this).val('');
});
db.php中:
<?php
$conn = mysql_connect('localhost','B00556019','73eKESV3') ;
$db = mysql_select_db('b00556019');
?>
userinfo.php:
<?php
include_once('db.php');
$name = $_POST['name'];
$age = $_POST['age'];
if(mysql_query("INSERT INTO user (name, age) VALUES (
'$name','$age')";))
echo "Successful";
else
echo "insertion failed";
?>
它不會發布到數據庫
數據庫名稱:b00556019
表:用戶
欄位:name(varchar,15)。 年齡(INT,3)
除了userinfo頁顯示為空白以外,沒有任何反應。
任何對如何發布到數據庫有任何建議的人都很好,並且如果有人有簡單的教程,因為這是基本的教程,但仍然行不通。
旁注:就目前而言,您可以進行SQL injection
請考慮切換到使用mysqli_*
函數以及已准備好的語句或PDO。 mysql_*
函數已被棄用,並將在以后的PHP版本中刪除。
代替:
if(mysql_query("INSERT INTO user (name, age) VALUES (
'$name','$age')";))
echo "Successful";
else
echo "insertion failed";
采用:
$name = mysql_real_escape_string($_POST['name']);
$age = mysql_real_escape_string($_POST['age']);
$result = mysql_query("INSERT INTO user (name, age) VALUES ('$name','$age')");
if(!mysql_query($result)){
echo "Insertion was not successful.";
} else {
echo "Insertion was successful.";
}
您可能還需要通過數據庫連接:
if(!mysql_query($result,$conn))
准備報表的方法:
$conn = new mysqli("xxx", "xxx", "xxx", "xxx");
/* check connection */
if (mysqli_connect_errno()) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
$name = $_POST['name'];
$age = $_POST['age'];
$result = $conn->prepare("INSERT INTO user (name, age) VALUES (?,?)");
$result->bind_param("ss", $name, $age);
$result->execute();
$result->close();
// If there is any error with your SQL statement an
// error is thrown and displayed to the user:
printf("Prepared Statement Error: %s\n", $conn->error);
首先更正表名稱:“ table:users”,它是您在代碼中使用的“ user”。
if(!$conn){
echo mysql_error() ; // first check while connecting..
}
// on the insertion code .....
if(mysql_query("INSERT INTO user (name, age) VALUES (
'$name','$age')";))
echo "Successful";
else
echo mysql_error($conn) ; // you will see what is the error..
echo "insertion failed";
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