Question

如果我想搜索文字而不是数字，我该如何更改我的代码？这是我的解析数据，当我输入数字时它运行但是当它的文本，我得到错误。谢谢你提前:)

            //parse json data
            try{

                JSONObject object = new JSONObject(result);
                String ch = object.getString("re");
                if (ch.equals("success")) {

                    JSONObject no = object.getJSONObject("0");

                    // long q=object.getLong("f1");
                    String dn = no.getString("deceased_name");
                    String c = no.getString("company");
                    String ca = no.getString("company_address");

                    editText1.setText(c);
                    editText2.setText(ca);
                    editText3.setText(dn);

                    Toast.makeText(getApplicationContext(),
                            "Born: " + String.valueOf(c), Toast.LENGTH_LONG)
                            .show();
                    Toast.makeText(getApplicationContext(),
                            "Died: " + String.valueOf(ca), Toast.LENGTH_LONG)
                            .show();
                    Toast.makeText(getApplicationContext(),
                            "Died: " + String.valueOf(dn), Toast.LENGTH_LONG)
                            .show();
                }

// php code

    <?php
     $con = mysql_connect("localhost","root","");
     if (!$con)
       {
         die('Could not connect: ' . mysql_error());
       }
       mysql_select_db("demo", $con);
       $first_name=$_REQUEST['first_name'];
       //$v1='111';

      if($first_name==NULL)
       {
            $r["re"]="Enter the number!!!";
             print(json_encode($r));
             die('Could not connect: ' . mysql_error());
       }
      else
       {     
        // $v1="530";      
       $i=mysql_query("select first_name from tblcontact where                               first_name=$first_name",$con);
       $check='';
      while($row = mysql_fetch_array($i))
        {
         $r[]=$row;
              $check=$row['first_name'];
             // print(json_encode($r));
         }
if($check==NULL)
       {            
                  $r["re"]="Record is not available";
                  print(json_encode($r));

         }
        else
         {
             $r["re"]="success";
             print(json_encode($r));
            // die('Could not connect: ' . mysql_error());                
          } 
      }
  mysql_close($con);             
?>

Answer 1

"select first_name from tblcontact where first_name = $first_name"

将此查询更改为

"select first_name from tblcontact where first_name = '". mysql_real_escape_string($first_name)."'"

JSONObject文本搜索到Mysql

1 个答案: