IDはphpとjquery、ajaxを使用して1ページから別のページに渡されていません

Question

私はデータベース内の異なるテーブルのすべてのレコード（タイトルのみ）を表示している検索バーを持っています。どのようにajaxを使ってあるページから別のページにidを渡すのですか？

Here is my code: 
ajax_search.php: 


<?php 
require_once('../config.php'); 
if($_GET['data']=='setLanguage') 
{ 
    $lang = $_POST['lang']; 
    $_SESSION['CHOOSEN_LANGUAGE'] = $lang; 
} 
// FUNCTION FOR INSERT CITY BASED ON LAT LONG // 
if($_GET['data']=='getCityByLatLong') 
{ 
    $browser_used=$_POST['browser_used']; 
    $city_id =''; 
    if($browser_used!='chrome') 
    { 
     $deal_lat=$_POST['lat']; 
     $deal_long=$_POST['lang']; 
     $geocode=file_get_contents('http://maps.googleapis.com/maps/api/geocode/json?latlng='.$deal_lat.','.$deal_long.'&sensor=false'); 
     $output= json_decode($geocode); 

     for($j=0;$j<count($output->results[0]->address_components);$j++) 
     { 
      $cn=array($output->results[0]->address_components[$j]->types[0]); 
      if(in_array("locality", $cn)) 
      { 
       $city_name= $output->results[0]->address_components[$j]->long_name; 
      } 
     } 
     $sql1 = "SELECT CITY_ID FROM CITY WHERE CITY_TITLE='$city_name'"; 
     $result1 = mysql_query($sql1); 
     $num = mysql_num_rows($result1); 
     if($num!='') 
     { 
      $city_row = mysql_fetch_assoc($result1); 
      $city_id = $city_row['CITY_ID']; 

     } 
    } 
    $city_id=100; 
    if($city_id !='') 
    { 
     $_SESSION['CHOOSEN_CITY_ID'] = $city_id; 
    } 
    echo $city_id; 
} 
if($_GET['data']=='changeCity') 
{ 
    $city_id = $_POST['city_id']; 
    echo $_SESSION['CHOOSEN_CITY_ID'] = $city_id; 
} 
//set SpecialityID in session 
if($_GET['data']=='setSpeciality') 
{ 
    unset($_SESSION['SPECIALITY']); 
    $speciality_id = $_POST['speciality_id']; 
    echo $_SESSION['SPECIALITY']['ID'] = $speciality_id; 
} 

if($_GET['data']=='getAreaByCity') 
{ 
    $city_id = $_POST['city_id']; 
    $sql1="SELECT CITY_AREA_ID,AREA_TITLE FROM CITY_AREA WHERE CITY_ID ='".$city_id."' AND AREA_STATUS ='1'"; 
    $response=mysql_query($sql1); 
    echo"<option value=''>Locality</option>"; 
    while($row = mysql_fetch_assoc($response)) 
    { 
     $id = $row['CITY_AREA_ID']; 
     $title = $row['AREA_TITLE']; 
     echo"<option value='$id'>$title</option>"; 
    }  
} 

//CREATE BY RAJAN *** Search Query*** 
//------------------------------------- 
//CREATE QUERY TO DB AND PUT RECEIVED DATA INTO ASSOCIATIVE ARRAY 
if (isset($_REQUEST['query'])) 
{ 
    $query = $_REQUEST['query']; 
    /*$querystr = "SELECT e.EMPLOYEE_ID,TRIM(IF('$LANG' = 'en',e.FIRST_NAME,e.FIRST_NAME_AR)) AS TITLE, 
    TRIM(IF('$LANG' = 'en',s.SPECIALIZATION_TITLE,s.SPECIALIZATION_AR_TITLE)) AS SPECIALIZATION_TITLE 
    FROM EMPLOYEE AS e 
    LEFT JOIN EMP_SPECIALIZATION AS sp ON sp.EMPLOYEE_ID=e.EMPLOYEE_ID 
    LEFT JOIN SPECIALIZATION AS s ON s.SPECIALIZATION_ID=sp.SPECIALIZATION_ID 
    LEFT JOIN EMPLOYEE_UNIT AS eu ON eu.EMPLOYEE_ID=e.EMPLOYEE_ID 
    LEFT JOIN UNIT AS u ON u.UNIT_ID=eu.UNIT_ID 
    LEFT JOIN UNIT_SERVICES AS us ON us.UNIT_ID=u.UNIT_ID 
    LEFT JOIN SERVICES AS sr ON sr.SERVICES_ID=us.SERVICES_ID 
    WHERE e.FIRST_NAME LIKE '%$query%' OR e.FIRST_NAME_AR LIKE '%$query%' OR e.LAST_NAME LIKE '%$query%' OR e.LAST_NAME_AR LIKE '%$query%' 
    OR s.SPECIALIZATION_TITLE LIKE '%$query%' OR s.SPECIALIZATION_AR_TITLE LIKE '%$query%' 
    OR u.UNIT_TITLE LIKE '%$query%' OR u.UNIT_AR_TITLE LIKE '%$query%' 
    OR sr.SERVICES_TITLE LIKE '%$query%' OR sr.SERVICES_AR_TITLE LIKE '%$query%' 
    GROUP BY EMPLOYEE_ID 
    ORDER BY TITLE"; 
    $sql = mysql_query ($querystr); 
    $array = array(); 

    while ($row = mysql_fetch_array($sql)) 
    { 
     $array[] = array ('label' => $row['EMPLOYEE_ID'],'value' => $row['TITLE'],); 
    }*/ 
    $array = array(); 
    $querystr = "SELECT e.EMPLOYEE_ID,TRIM(CONCAT(IF('$LANG' = 'en',e.FIRST_NAME,e.FIRST_NAME_AR),IF('$LANG' = 'en',e.LAST_NAME,e.LAST_NAME_AR))) AS TITLE, 
    TRIM(IF('$LANG' = 'en',s.SPECIALIZATION_TITLE,s.SPECIALIZATION_AR_TITLE)) AS DEPARTMENT 
    FROM EMPLOYEE AS e 
    LEFT JOIN EMP_SPECIALIZATION AS sp ON sp.EMPLOYEE_ID=e.EMPLOYEE_ID 
    LEFT JOIN SPECIALIZATION AS s ON s.SPECIALIZATION_ID=sp.SPECIALIZATION_ID 
    WHERE e.FIRST_NAME LIKE '%$query%' OR e.FIRST_NAME_AR LIKE '%$query%' OR e.LAST_NAME LIKE '%$query%' OR e.LAST_NAME_AR LIKE '%$query%' 
    OR s.SPECIALIZATION_TITLE LIKE '%$query%' OR s.SPECIALIZATION_AR_TITLE LIKE '%$query%' 
    GROUP BY EMPLOYEE_ID 
    ORDER BY TITLE"; 
    $sql = mysql_query ($querystr); 

    $cnt = mysql_num_rows($sql); 
    while ($row = mysql_fetch_array($sql)) 
    { 
     $array[] = array ('value' => $row['TITLE']." --From Specilization"); 
    } 

    $querystr = "SELECT e.EMPLOYEE_ID,TRIM(CONCAT(IF('$LANG' = 'en',e.FIRST_NAME,e.FIRST_NAME_AR),IF('$LANG' = 'en',e.LAST_NAME,e.LAST_NAME_AR))) AS TITLE, 
    TRIM(IF('$LANG' = 'en',u.UNIT_TITLE,u.UNIT_AR_TITLE)) AS DEPARTMENT 
    FROM EMPLOYEE AS e 
    LEFT JOIN EMPLOYEE_UNIT AS eu ON eu.EMPLOYEE_ID=e.EMPLOYEE_ID 
    LEFT JOIN UNIT AS u ON u.UNIT_ID=eu.UNIT_ID 
    WHERE e.FIRST_NAME LIKE '%$query%' OR e.FIRST_NAME_AR LIKE '%$query%' OR e.LAST_NAME LIKE '%$query%' OR e.LAST_NAME_AR LIKE '%$query%' 
    OR u.UNIT_TITLE LIKE '%$query%' OR u.UNIT_AR_TITLE LIKE '%$query%' 
    GROUP BY EMPLOYEE_ID 
    ORDER BY TITLE"; 
    $sql = mysql_query ($querystr); 

    $cnt = mysql_num_rows($sql); 
    while ($row = mysql_fetch_array($sql)) 
    { 
     $array[] = array ('value' => $row['TITLE']." --From Unit"); 
    } 

    $querystr = "SELECT e.EMPLOYEE_ID,TRIM(CONCAT(IF('$LANG' = 'en',e.FIRST_NAME,e.FIRST_NAME_AR),IF('$LANG' = 'en',e.LAST_NAME,e.LAST_NAME_AR))) AS TITLE, 
    TRIM(IF('$LANG' = 'en',sr.SERVICES_TITLE,sr.SERVICES_AR_TITLE)) AS DEPARTMENT 
    FROM EMPLOYEE AS e 
    LEFT JOIN EMPLOYEE_UNIT AS eu ON eu.EMPLOYEE_ID=e.EMPLOYEE_ID 
    LEFT JOIN UNIT AS u ON u.UNIT_ID=eu.UNIT_ID 
    LEFT JOIN UNIT_SERVICES AS us ON us.UNIT_ID=u.UNIT_ID 
    LEFT JOIN SERVICES AS sr ON sr.SERVICES_ID=us.SERVICES_ID 
    WHERE e.FIRST_NAME LIKE '%$query%' OR e.FIRST_NAME_AR LIKE '%$query%' OR e.LAST_NAME LIKE '%$query%' OR e.LAST_NAME_AR LIKE '%$query%' 
    OR sr.SERVICES_TITLE LIKE '%$query%' OR sr.SERVICES_AR_TITLE LIKE '%$query%' 
    GROUP BY EMPLOYEE_ID 
    ORDER BY TITLE"; 
    $sql = mysql_query ($querystr); 

    $cnt = mysql_num_rows($sql); 
    while ($row = mysql_fetch_array($sql)) 
    { 
     $array[] = array ('value' => $row['TITLE']." --From Services ".$row['EMPLOYEE_ID']); 
    } 

    $querystr = "SELECT u.UNIT_ID,TRIM(IF('$LANG' = 'en',u.UNIT_TITLE,u.UNIT_AR_TITLE)) AS TITLE 
    FROM UNIT AS u 
    WHERE u.UNIT_TITLE LIKE '%$query%' OR u.UNIT_AR_TITLE LIKE '%$query%' 
    GROUP BY UNIT_ID 
    ORDER BY TITLE"; 
    $sql = mysql_query ($querystr); 
    $cnt = mysql_num_rows($sql); 
    while ($row = mysql_fetch_array($sql)) 
    { 
     $array[] = array ('value' => $row['TITLE']." --From Clinic"); 
    } 

    //RETURN JSON ARRAY 
    echo json_encode ($array); 
} 
?> 

search_section.php: 

<div class="form-group check1"> 
         <label class="sr-only" for="location">Search here for Doctor's Name, Specialities or Hospitals</label> 
         <input type="hidden" name="test_id" id="test_id" value=""> 
         <input type="text" name="search_specialist" size="30" onchange="return getDetails();" class="form-control newcheck" id="location" placeholder="Search here for Doctor's Name, Specialities or Hospitals"> 
         <button type="submit" class="btn btn-default btn-primary prim" onclick="return getDetails();"><i class="fa fa-search" aria-hidden="true"></i></button> 
        </div> 

<script> 
    $('input.newcheck').typeahead({ 
     name: 'search_specialist', 
     remote:"<?php echo CUR_URL ;?>new_template/ajax_code.php?query=%QUERY", 
     limit: 100 
</script> 

Answer 1

病気Okejは、単にAJAXを使用して値を渡す方法を、あなたの質問に答えるためにしようと...私はこの問題を解決するために助けてください。フォームで

<form id="form"> 
<input type="text" id="input" value="test"> 
</form> 

すべての入力フィールドの値が渡されます： まずあなたがフォームを必要としています。

アヤックス：あなたはまだ表示ブロック、例での入力を追加することができます

jQuery(function(){ 
$('#form').on('submit', function (e) { //on submit function 
    e.preventDefault(); 
    $.ajax({ 
     type: 'post', //method POST 
     url: 'yoururl.php', //URL of page where to pass values 
     data: $('#form').serialize(), //seriallize is passing all inputs values of form 
     success: function(){ //on success function 
       $("#input").attr("disabled", true); //example 
       $("#input").removeClass('btn-primary').addClass('btn-success');//example 
        }, 
    }); 
} 
}); 

注意（uはuは、ユーザーがページに表示したくないいくつかの値を渡す場合）：

<input type="text" name="akcija" value="<?php echo $row['akcija']; ?>" style="display:none;" /> 

そして、ajaxページでは、値を取得することができます

$akcija = $_POST['akcija']