eclipse android使用HTTP将java值传递给php(eclipse android pass a java value to php by using HTTP)

eclipse android使用HTTP将java值传递给php(eclipse android pass a java value to php by using HTTP)

我想将变量q从java传递给php。

String q = "select author from books where 1"; try{ HttpClient httpClient = new DefaultHttpClient(); HttpPost httpPost = new HttpPost("http://www2.XXXX./XXXX/X.php?qy="+q); //"http://10.0.2.2/tut.php", http://www.XXXX/XXXX/tut.php HttpRespe respe = (httpPost); HttpEntity entity = respe.getEntity(); is = entity.getContent(); }catch(Exception e){ e.printStackTrace(); println("Exception 1 caught"); }

但是，php文件无法从java（php正确连接到mysql）获取值。

php编码：

<?php $con=mysql_connect("XXX.XXX","XX","XXX"); mysql_select_db("XX",$con); $st = $_GET['qy']; $r = mysql_query("$st"); while($row=mysql_fetch_array($r)) { $out[]=$row; } print(json_encode($out)); mysql_close($con); ?>

我发现如果我只是将表名传递给php，它就可以了。 但是如果传递的变量变得更长，它就会陷入困境。 我怎样才能解决这个问题？ 如何将多个变量传递给php（即mysql_query("select $_GET['col'] from $_GET['table'] where $_GET['condition']"); ）？

I want to pass the variable q from java to php.

String q = "select author from books where 1"; try{ HttpClient httpClient = new DefaultHttpClient(); HttpPost httpPost = new HttpPost("http://www2.XXXX./XXXX/X.php?qy="+q); //"http://10.0.2.2/tut.php", http://www.XXXX/XXXX/tut.php HttpRespe respe = (httpPost); HttpEntity entity = respe.getEntity(); is = entity.getContent(); }catch(Exception e){ e.printStackTrace(); println("Exception 1 caught"); }

However, the php file cannot get the value from java(php connected to mysql correctly).

php coding:

<?php $con=mysql_connect("XXX.XXX","XX","XXX"); mysql_select_db("XX",$con); $st = $_GET['qy']; $r = mysql_query("$st"); while($row=mysql_fetch_array($r)) { $out[]=$row; } print(json_encode($out)); mysql_close($con); ?>

I found that if I just pass the table name to php, it works. But if the passing variable become longer, it went to caught one. How can I fix this? How about passing more than one variable to php ( mysql_query("select $_GET['col'] from $_GET['table'] where $_GET['condition']");)?

最满意答案

使用Post而不是Get，它更安全。

像这样的东西：

String col = "author"; String table = "books"; String condition = "1"; try{ List<ameValuePair> params = new ArrayList<ameValuePair>(); params.add(new BasicameValuePair("col", col)); params.add(new BasicameValuePair("table", table)); params.add(new BasicameValuePair("condition", condition)); HttpClient httpClient = new DefaultHttpClient(); HttpPost httpPost = new HttpPost("http://www2.XXXX./XXXX/X.php"); httpPost.setEntity(new UrlEncodedFormEntity(params)); HttpRespe respe = (httpPost); HttpEntity entity = respe.getEntity(); is = entity.getContent(); }catch(Exception e){ e.printStackTrace(); println("Exception 1 caught"); }

PHP：

<?php if (isset($_POST['col']) && isset($_POST['table']) && isset($_POST['condition'])){ $columname= $_POST['col']; $tableame = $_POST['table']; $condition = $_POST['condition']; $dbh=mysql_connect ("localhost", "username", "password") or die('Cannot connect to the database because: ' . mysql_error()); mysql_select_db ("database_name"); $sql=mysql_query("select '$columname' from '$tableame' where '$condition'"); while($row=mysql_fetch_assoc($sql)) $output[]=$row; print(json_encode($output)); mysql_close(); } ?>

它工作（由mandi yeung）

Use Post instead of Get, it's more secure.

Something like this:

String col = "author"; String table = "books"; String condition = "1"; try{ List<ameValuePair> params = new ArrayList<ameValuePair>(); params.add(new BasicameValuePair("col", col)); params.add(new BasicameValuePair("table", table)); params.add(new BasicameValuePair("condition", condition)); HttpClient httpClient = new DefaultHttpClient(); HttpPost httpPost = new HttpPost("http://www2.XXXX./XXXX/X.php"); httpPost.setEntity(new UrlEncodedFormEntity(params)); HttpRespe respe = (httpPost); HttpEntity entity = respe.getEntity(); is = entity.getContent(); }catch(Exception e){ e.printStackTrace(); println("Exception 1 caught"); }

PHP:

<?php if (isset($_POST['col']) && isset($_POST['table']) && isset($_POST['condition'])){ $columname= $_POST['col']; $tableame = $_POST['table']; $condition = $_POST['condition']; $dbh=mysql_connect ("localhost", "username", "password") or die('Cannot connect to the database because: ' . mysql_error()); mysql_select_db ("database_name"); $sql=mysql_query("select '$columname' from '$tableame' where '$condition'"); while($row=mysql_fetch_assoc($sql)) $output[]=$row; print(json_encode($output)); mysql_close(); } ?>

it works(by mandi yeung)