尝试在数据库GET方法上插入时发生PHP错误
当我试图在数据库中插入两个值时,我的PHP有问题 我正在使用Android,我调用php进行插入。我发布我的Android代码:尝试在数据库GET方法上插入时发生PHP错误,php,android,mysql,get,Php,Android,Mysql,Get,当我试图在数据库中插入两个值时,我的PHP有问题 我正在使用Android,我调用php进行插入。我发布我的Android代码: private void loadPhase() { String url = "http://www.myurl.com/scriptsfolder/myphptoinsertranking.php?id=prueba0101&points=200"; if (numQuestion < 11) { header4q
private void loadPhase() {
String url = "http://www.myurl.com/scriptsfolder/myphptoinsertranking.php?id=prueba0101&points=200";
if (numQuestion < 11) {
header4questions.setText(listQuestions.get(questionPosition));
answer1.setText(listAnswers.get(questionPosition * 4));
answer2.setText(listAnswers.get((questionPosition * 4) + 1));
answer3.setText(listAnswers.get((questionPosition * 4) + 2));
answer4.setText(listAnswers.get((questionPosition * 4) + 3));
if(numQuestion % 5 == 0 && numQuestion > 0)
{
MediaPlayer mp = MediaPlayer.create(this, R.raw.aplausos);
mp.start();
}
numQuestion++;
}
else {
Toast.makeText(getApplicationContext(), "You have finished the quiz!", Toast.LENGTH_SHORT).show();
StringRequest stringRequest = new StringRequest(Request.Method.GET, url, new Response.Listener<String>() {
@Override
public void onResponse(String response) {
Toast.makeText(QuizActivity.this,response,Toast.LENGTH_LONG).show();
Intent inn1 = getIntent();
inn1 = new Intent(QuizActivity.this, MainActivity.class);
startActivity(inn1);
}
},
new Response.ErrorListener() {
@Override
public void onErrorResponse(VolleyError error) {
Toast.makeText(QuizActivity.this, "INSERT ERROR", Toast.LENGTH_LONG).show();
Intent inn1 = getIntent();
inn1 = new Intent(QuizActivity.this, MainActivity.class);
startActivity(inn1);
}
}) {
};
//Adding the string request to the queue
RequestQueue requestQueue = Volley.newRequestQueue(this);
requestQueue.add(stringRequest);
}
}
private void loadPhase(){
字符串url=”http://www.myurl.com/scriptsfolder/myphptoinsertranking.php?id=prueba0101&points=200";
if(numQuestion<11){
header4questions.setText(listQuestions.get(questionPosition));
答案1.setText(listAnswers.get(问题位置*4));
答案2.setText(listAnswers.get((问题位置*4)+1));
答案3.setText(listAnswers.get((问题位置*4)+2));
答案4.setText(listAnswers.get((问题位置*4)+3));
如果(numQuestion%5==0&&numQuestion>0)
{
MediaPlayer mp=MediaPlayer.create(this,R.raw.aplausos);
mp.start();
}
numQuestion++;
}
否则{
Toast.makeText(getApplicationContext(),“您已经完成了测试!”,Toast.LENGTH\u SHORT.show();
StringRequest StringRequest=新建StringRequest(Request.Method.GET,url,new Response.Listener()){
@凌驾
公共void onResponse(字符串响应){
Toast.makeText(quizaActivity.this,response,Toast.LENGTH_LONG.show();
Intent inn1=getIntent();
inn1=新意图(quizaActivity.this,MainActivity.class);
星触觉(inn1);
}
},
新的Response.ErrorListener(){
@凌驾
公共无效onErrorResponse(截击错误){
Toast.makeText(QuizActivity.this,“插入错误”,Toast.LENGTH_LONG.show();
Intent inn1=getIntent();
inn1=新意图(quizaActivity.this,MainActivity.class);
星触觉(inn1);
}
}) {
};
//将字符串请求添加到队列
RequestQueue RequestQueue=Volley.newRequestQueue(this);
添加(stringRequest);
}
}
我还发布了我的php代码:
<?php
if($_SERVER['REQUEST_METHOD']=='GET'){
$id = $_POST['id'];
$points = $_POST['points'];
$con = mysqli_connect("server.com:1111","username","userpassword","databasename");
if (!$con->connect_error)
{
$sql= "INSERT INTO table (IdUser, Points) values ('$id','$points')";
if(mysqli_query($con,$sql)){
echo "Successfully Registered";
}
else{
echo "Could not register";
}
}
else{
echo 'error de conexion';
}
}
?>
它应该是$\u GET
它应该是$\u GET
您在检查POST时发出GET请求,请将php的代码更改为
<?php
if($_SERVER['REQUEST_METHOD']=='GET'){
$id = $_GET['id'];
$points = $_GET['points'];
$con = mysqli_connect("server.com:1111","username","userpassword","databasename");
if (!$con->connect_error)
{
$sql= "INSERT INTO table (IdUser, Points) values ('$id','$points')";
if(mysqli_query($con,$sql)){
echo "Successfully Registered";
}
else{
echo "Could not register";
}
}
else{
echo 'error de conexion';
}
}
?>
您在检查POST时发出GET请求,请将php代码更改为
<?php
if($_SERVER['REQUEST_METHOD']=='GET'){
$id = $_GET['id'];
$points = $_GET['points'];
$con = mysqli_connect("server.com:1111","username","userpassword","databasename");
if (!$con->connect_error)
{
$sql= "INSERT INTO table (IdUser, Points) values ('$id','$points')";
if(mysqli_query($con,$sql)){
echo "Successfully Registered";
}
else{
echo "Could not register";
}
}
else{
echo 'error de conexion';
}
}
?>
更改此选项
$id=$_POST['id'];
$points=$_POST['points']
对此,
$id=$_GET['id'];
$points=$_获得['points'] 改变这个
$id=$_POST['id'];
$points=$_POST['points']
对此,
$id=$_GET['id'];
$points=$_获得['points'] 使用$\u GET
代替$\u POST
使用$\u GET
代替$\u POST
好的,我会试试。你认为还有什么问题吗?这可以解决它吗?直到现在我看到这个错误,希望这能解决所有问题,我会尝试。你认为还有什么问题吗?这可以解决它吗?直到现在我看到这个错误,希望这能解决所有问题,我会尝试。你觉得还有什么问题吗?这个可以解决吗?好的,我试试。你觉得还有什么问题吗?这个可以解决吗?好的,我试试。你觉得还有什么问题吗?这个可以解决吗?好的,我试试。你觉得还有什么不对劲吗?这能解决问题吗?