[英]PHP not return expected result
我得到一個非常奇怪的結果! 我從Java類發布了一個id
,其中該ID將在php腳本中用於檢索特定數據。 該value
應為1,但始終顯示2
<?php
if($_SERVER['REQUEST_METHOD']=='POST'){
//Getting values
$id = $_POST['id'];
//Creating sql query
$sql = "SELECT xuenian FROM student WHERE sid='$id'";
//importing dbConnect.php script
require_once('db_config.php');
//executing query
$result = mysqli_query($con,$sql);
$value = mysqli_fetch_object($result);
$value->xuenian;
if($value === "1"){
echo "1";
}else{
echo "2";
}
mysqli_close($con);
}
我試過==
,結果還是一樣。
Java類
public void loadResults(final String id, final int xuenian) {
StringRequest stringRequest = new StringRequest(Request.Method.POST, AppConfig.URL_CHECKID,
new Response.Listener<String>() {
@Override
public void onResponse(String response) {
Toast.makeText(getApplication(),response+"from php",Toast.LENGTH_LONG).show();
}
},
new Response.ErrorListener() {
@Override
public void onErrorResponse(VolleyError error) {
Toast.makeText(getApplication(), error + "", Toast.LENGTH_LONG).show();
}
}) {
@Override
protected Map<String, String> getParams() throws AuthFailureError {
Map<String, String> params = new HashMap<>();
//Adding parameters to request
params.put(AppConfig.KEY_USERID, id);
//returning parameter
return params;
}
};
//Adding the string request to the queue
RequestQueue requestQueue = Volley.newRequestQueue(this);
requestQueue.add(stringRequest);
}
您正在此處將$value
設置$value
對象:
$value = mysqli_fetch_object($result);
然后此行不執行任何操作:
$value->xuenian;
在下一行, $value
仍然是一個對象,但是您將其與字符串進行比較,該字符串將始終為false:
if($value === "1")
{
echo "1";
}else{
echo "2";
}
您可能想要這樣:
if($value->xuenian === "1")
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.