I'm trying to send a string and returns json array of objects in response like like this:
我正在尝试发送一个字符串并返回响应的json数组,如下所示:
// Creating volley request obj
JsonArrayRequest movieReq = new JsonArrayRequest(url,
new Response.Listener<JSONArray>() {
@Override
public void onResponse(JSONArray response) {
Log.d(TAG, response.toString());
hidePDialog();
// Parsing json
for (int i = 0; i < response.length(); i++) {
try {
JSONObject obj = response.getJSONObject(i);
Movie movie = new Movie();
movie.setTitle(obj.optString("fullname"));
movie.setThumbnailUrl(obj.optString("image"));
movie.setRating(obj.optString("location"));
movie.setYear(obj.getInt("id"));
// adding movie to movies array
movieList.add(movie);
} catch (JSONException e) {
e.printStackTrace();
}
}
// notifying list adapter about data changes
// so that it renders the list view with updated data
// adapter.notifyDataSetChanged();
adapter.reloadData();
}
}, new Response.ErrorListener() {
@Override
public void onErrorResponse(VolleyError error) {
VolleyLog.d(TAG, "Error: " + error.getMessage());
hidePDialog();
}
})
{
@Override
protected Map<String, String> getParams() {
Map<String, String> params = new HashMap<String, String>();
params.put("fullname", "test"); // fullname is variable and test is value.
return params;
} };
In above code fullname is variable and test is value and using that variable I'm trying to send my variable data in my php variable and perform my query like this:
在上面的代码中,fullname是变量,test是值,并且使用该变量我正在尝试在我的php变量中发送我的变量数据并执行我的查询,如下所示:
<?php
header("content-type:application/json");
require_once("dbConnect.php");
$fullname = $_REQUEST["fullname"];
//echo $fullname."11";
$sql = "SELECT id ,image,fullname,location from uploadfinding WHERE fullname like '%$fullname%'";
$res = mysqli_query($conn,$sql);
$result = array();
while($row = mysqli_fetch_array($res)){
array_push($result, array(
"id"=>$row["id"],
"fullname"=>$row["fullname"],
"image"=>$row['image'],
"location"=>$row["location"]));
//echo " over";
}
$fp = fopen('results.json', 'w');
fwrite($fp, json_encode($result));
fclose($fp);
echo json_encode($result);
mysqli_close($conn);
?>
But my value test not transfer in php variable $fullname. So problem is how to transfer value test to php variable $fullname.
但是我的值测试没有在php变量$ fullname中传输。所以问题是如何将值测试转移到php变量$ fullname。
1 个解决方案
#1
0
If you want to receive test data the in PHP script then use this code. You make sure your request method type.
如果要在PHP脚本中接收测试数据,请使用此代码。您确保您的请求方法类型。
<?php
header("content-type:application/json");
require_once("dbConnect.php");
if($_SERVER['REQUEST_METHOD']=='POST')
{
$fullname = $_POST["fullname"];
echo "Received fullname = ".$fullname;
}
else if($_SERVER['REQUEST_METHOD']=='GET')
{
$fullname = $_POST["fullname"];
echo "Received fullname = ".$fullname;
}
$sql = "SELECT id ,image,fullname,location from uploadfinding WHERE fullname like '%$fullname%'";
$res = mysqli_query($conn,$sql);
if($res)
{
$result = array();
$result["detial"] = array();
while($row = mysqli_fetch_array($res)){
// temporary array to create single category
$tmp = array();
$tmp["id"] = $row["id"];
$tmp["fullname"] = $row["fullname"];
$tmp["image"] = $row["image"];
$tmp["location"] = $row["location"];
$tmp["des"] = $row["ProductDescription"];
// push category to final json array
array_push($result["detial"], $tmp);
}
// keeping response header to json
header('Content-Type: application/json');
// echoing json result
echo json_encode($result);
}
else
{
echo "No date found ";
}
?>
For more detail please check this tutorial and read this and this documentation.
有关更多详细信息,请查看本教程并阅读本文档和本文档。
I hope this will help you and you get your answer.
我希望这会对你有帮助,你得到答案。
#1
0
If you want to receive test data the in PHP script then use this code. You make sure your request method type.
如果要在PHP脚本中接收测试数据,请使用此代码。您确保您的请求方法类型。
<?php
header("content-type:application/json");
require_once("dbConnect.php");
if($_SERVER['REQUEST_METHOD']=='POST')
{
$fullname = $_POST["fullname"];
echo "Received fullname = ".$fullname;
}
else if($_SERVER['REQUEST_METHOD']=='GET')
{
$fullname = $_POST["fullname"];
echo "Received fullname = ".$fullname;
}
$sql = "SELECT id ,image,fullname,location from uploadfinding WHERE fullname like '%$fullname%'";
$res = mysqli_query($conn,$sql);
if($res)
{
$result = array();
$result["detial"] = array();
while($row = mysqli_fetch_array($res)){
// temporary array to create single category
$tmp = array();
$tmp["id"] = $row["id"];
$tmp["fullname"] = $row["fullname"];
$tmp["image"] = $row["image"];
$tmp["location"] = $row["location"];
$tmp["des"] = $row["ProductDescription"];
// push category to final json array
array_push($result["detial"], $tmp);
}
// keeping response header to json
header('Content-Type: application/json');
// echoing json result
echo json_encode($result);
}
else
{
echo "No date found ";
}
?>
For more detail please check this tutorial and read this and this documentation.
有关更多详细信息,请查看本教程并阅读本文档和本文档。
I hope this will help you and you get your answer.
我希望这会对你有帮助,你得到答案。