Php 使用where子句检索MySQL数据
我想知道如何使用Php 使用where子句检索MySQL数据,php,android,mysql,listview,where-clause,Php,Android,Mysql,Listview,Where Clause,我想知道如何使用WHERE子句从MySQL获取数据,并最终加载到androidlistView?我想根据姓名和月份获取日期、时间和超时。这就是我迄今为止所尝试的 获取数据 public void getData(String name, String month) { class GetDataJSON extends AsyncTask<String, Void, String> { @Override prote
WHERE
子句从MySQL
获取数据,并最终加载到androidlistView
?我想根据姓名和月份获取日期、时间和超时。这就是我迄今为止所尝试的
获取数据
public void getData(String name, String month) {
class GetDataJSON extends AsyncTask<String, Void, String> {
@Override
protected String doInBackground(String... params) {
DefaultHttpClient httpclient = new DefaultHttpClient(new BasicHttpParams());
HttpPost httppost = new HttpPost("http://192.168.1.7/Android/CRUD/retrieveInformation.php");
// Depends on your web service
httppost.setHeader("Content-type", "application/json");
InputStream inputStream = null;
String result = null;
try {
HttpResponse response = httpclient.execute(httppost);
HttpEntity entity = response.getEntity();
inputStream = entity.getContent();
// json is UTF-8 by default
BufferedReader reader = new BufferedReader(new InputStreamReader(inputStream, "UTF-8"), 8);
StringBuilder sb = new StringBuilder();
String line = null;
while ((line = reader.readLine()) != null)
{
sb.append(line + "\n");
}
result = sb.toString();
} catch (Exception e) {
// Oops
}
finally {
try{if(inputStream != null)inputStream.close();}catch(Exception squish){}
}
return result;
}
@Override
protected void onPostExecute(String result){
myJSON=result;
showList();
}
}
GetDataJSON g = new GetDataJSON();
g.execute();
}
protected void showList(){
try {
JSONObject jsonObj = new JSONObject(myJSON);
information = jsonObj.getJSONArray(Config.TAG_RESULTS);
for(int i=0;i<information.length();i++){
JSONObject c = information.getJSONObject(i);
String date = c.getString(Config.TAG_DATE);
String timeIn = c.getString(Config.TAG_TiME_IN);
String timeOut = c.getString(Config.TAG_TIME_OUT);
HashMap<String,String> info = new HashMap<String,String>();
info.put(Config.TAG_DATE, date);
info.put(Config.TAG_TiME_IN, timeIn);
info.put(Config.TAG_TIME_OUT,timeOut);
infoList.add(info);
}
ListAdapter adapter = new SimpleAdapter(
HomePage.this, infoList, R.layout.retrieve_data,
new String[]{Config.TAG_DATE,Config.TAG_TiME_IN,Config.TAG_TIME_OUT},
new int[]{R.id.date,R.id.timeIn,R.id.timeOut}
);
listView.setAdapter(adapter);
} catch (JSONException e) {
e.printStackTrace();
}
}
public void getData(字符串名称、字符串月份){
类GetDataJSON扩展了AsyncTask{
@凌驾
受保护的字符串doInBackground(字符串…参数){
DefaultHttpClient httpclient=新的DefaultHttpClient(新的BasicHttpParams());
HttpPost HttpPost=新的HttpPost(“http://192.168.1.7/Android/CRUD/retrieveInformation.php");
//取决于您的web服务
setHeader(“内容类型”、“应用程序/json”);
InputStream InputStream=null;
字符串结果=null;
试一试{
HttpResponse response=httpclient.execute(httppost);
HttpEntity=response.getEntity();
inputStream=entity.getContent();
//默认情况下,json是UTF-8
BufferedReader=新的BufferedReader(新的InputStreamReader(inputStream,“UTF-8”),8);
StringBuilder sb=新的StringBuilder();
字符串行=null;
而((line=reader.readLine())!=null)
{
sb.追加(第+行“\n”);
}
结果=sb.toString();
}捕获(例外e){
//哎呀
}
最后{
尝试{if(inputStream!=null)inputStream.close();}捕获(异常挤压){}
}
返回结果;
}
@凌驾
受保护的void onPostExecute(字符串结果){
myJSON=结果;
showList();
}
}
GetDataJSON g=新的GetDataJSON();
g、 执行();
}
受保护的无效显示列表(){
试一试{
JSONObject jsonObj=新的JSONObject(myJSON);
information=jsonObj.getJSONArray(Config.TAG_RESULTS);
对于(inti=0;i您没有在HTTP请求中输入名称和月份
试试这个
HttpPost httppost = new HttpPost("http://192.168.1.7/Android/CRUD/retrieveInformation.php?name="+name+"&month="+month);
希望这有帮助:)您没有在HTTP请求中输入姓名和月份
试试这个
HttpPost httppost = new HttpPost("http://192.168.1.7/Android/CRUD/retrieveInformation.php?name="+name+"&month="+month);
希望这有帮助:)看起来您希望将name/month作为参数传递给PHP文件。
最简单的方法是在URL字符串中传递参数(最简单的方法):
然后,name和month被视为PHP文件中的$\u GET变量
但是,因为您希望您的代码被正确编码,所以您可以这样做:
URI uri = new URIBuilder()
.setScheme("http")
.setHost("192.168.1.7")
.setPath("/Android/CRUD/retrieveInformation.php")
.addParameter("name", name)
.addParameter("month", month)
.build();
HttpGet httpget = new HttpGet( uri );
请注意,我在第二个示例中使用的是HttpGet
,而不是HttpPost
关于PHP代码的快速建议,这样您就不必将所有索引重新映射到它们的键名:
while($row=mysqli_fetch_assoc($res)){
$result[] = $row;
}
mysqli\u fetch\u assoc将数组的键设置为列名。这将发送所有列。如果不想发送所有列,而只发送7列,则应将select查询修改为:
$sql = "select id, name, weather, date, status, time_in, time_out from information WHERE name= '". $name."' and month = '".$month."'";
并且,在选择查询之前,还要清理$name和$month:
$name = mysqli_escape_string($name);
$month = mysqli_escape_string($month);
以下是更新后的代码,以反映修改:
<?php
define('HOST','127.0.0.1:3307');
define('USER','root');
define('PASS','');
define('DB','androiddb');
$con = mysqli_connect(HOST,USER,PASS,DB) or die('unable to connect');
$name = $_GET['name'];
$month = $_GET['month'];
$months = [
'January' => '01',
'February' => '02',
'March' => '03',
'April' => '04',
'May' => '05',
'June' => '06',
'July' => '07',
'August' => '08',
'September' => '09',
'October' => '10',
'November' => '11',
'December' => '12',
];
if (!isset($months[ $month ])) {
die("Invalid month");
}
$month = $months[ $month ];
$name = mysqli_escape_string($con, $name);
$month = mysqli_escape_string($con, $month);
$sql = "select * from information WHERE name= '". $name."' and month = '".$month."'";
$res = mysqli_query($con,$sql);
$result=array();
while($row=mysqli_fetch_assoc($res)){
$result[] = $row;
}
echo (json_encode(array("result"=>$result)));
mysqli_close($con);
?>
看起来您希望将name/month作为参数传递给PHP文件。
最简单的方法是在URL字符串中传递参数(最简单的方法):
然后,name和month被视为PHP文件中的$\u GET变量
但是,因为您希望您的代码被正确编码,所以您可以这样做:
URI uri = new URIBuilder()
.setScheme("http")
.setHost("192.168.1.7")
.setPath("/Android/CRUD/retrieveInformation.php")
.addParameter("name", name)
.addParameter("month", month)
.build();
HttpGet httpget = new HttpGet( uri );
请注意,我在第二个示例中使用的是HttpGet
,而不是HttpPost
关于PHP代码的快速建议,这样您就不必将所有索引重新映射到它们的键名:
while($row=mysqli_fetch_assoc($res)){
$result[] = $row;
}
mysqli\u fetch\u assoc将数组的键设置为列名。这将发送所有列。如果不想发送所有列,而只发送7列,则应将select查询修改为:
$sql = "select id, name, weather, date, status, time_in, time_out from information WHERE name= '". $name."' and month = '".$month."'";
并且,在选择查询之前,还要清理$name和$month:
$name = mysqli_escape_string($name);
$month = mysqli_escape_string($month);
以下是更新后的代码,以反映修改:
<?php
define('HOST','127.0.0.1:3307');
define('USER','root');
define('PASS','');
define('DB','androiddb');
$con = mysqli_connect(HOST,USER,PASS,DB) or die('unable to connect');
$name = $_GET['name'];
$month = $_GET['month'];
$months = [
'January' => '01',
'February' => '02',
'March' => '03',
'April' => '04',
'May' => '05',
'June' => '06',
'July' => '07',
'August' => '08',
'September' => '09',
'October' => '10',
'November' => '11',
'December' => '12',
];
if (!isset($months[ $month ])) {
die("Invalid month");
}
$month = $months[ $month ];
$name = mysqli_escape_string($con, $name);
$month = mysqli_escape_string($con, $month);
$sql = "select * from information WHERE name= '". $name."' and month = '".$month."'";
$res = mysqli_query($con,$sql);
$result=array();
while($row=mysqli_fetch_assoc($res)){
$result[] = $row;
}
echo (json_encode(array("result"=>$result)));
mysqli_close($con);
?>
我需要在getData
中更改/添加任何内容吗?我不这么认为。如果您更改了URL,它可能会起作用。@John Joe:Ye Min Htut(正确!)说,如果您要在PHP中执行$\u GET['name'];
,那么您最好在URL中有一个参数name=xyz
。与“month”相同:这两个参数都需要在URL中指定。另外:看看Clayton的回答。他说的是同一件事。Clayton还(正确!)指出您可能希望使用URIBuilder
,您可以使用HttpGet()
而不是“HttpPost()”。是的。这是正确的,但您应该清理您的输入(姓名和月份)在sql查询中用作变量以保护sql注入。谢谢,将去看看我是否需要在getData
中更改/添加任何内容?我不这么认为。如果您更改了URL,它可能会起作用。@John Joe:Ye Min Htut(正确!)说如果您要执行$\u GET['name'];
在你的PHP中……那么你最好在你的URL中有一个参数name=xyz
。与“月”相同:两个参数都需要在URL中指定。另外:看看克莱顿的回答。他说的是同样的话。克莱顿也(正确!)指出您可能希望使用URIBuilder
,并且可以使用HttpGet()
而不是“HttpPost()”。是的。这是正确的,但您应该清理在sql查询中用作变量的输入(名称和月份),以保护sql注入。谢谢,将去看看您是否需要