鏈接下拉選項不通過拉入先前的值來查詢

我找到了一個鏈接下拉框的例子，但我努力確保第三個框在顯示所需結果時使用前面的框。鏈接下拉選項不通過拉入先前的值來查詢

的方框是爲了顯示下面，

下拉框1 =選擇一個扇區

下拉框2 =選擇一個級別

下拉框3 =選擇一個資格

在最後一個SQL查詢顯示$ HELP，我認爲這是我有我的問題，我認爲這是丟失在下拉框2選擇時，在下拉框1中以前存儲的值。

<?php 
//************************************** 
//  Page load dropdown results  // 
//************************************** 
function getTierOne() 
{ 
    $result = mysql_query("SELECT DISTINCT SSA1Text FROM qualifications ORDER BY SSA1Text ASC") 
    or die(mysql_error()); 

     while($tier = mysql_fetch_array($result)) 

     { 
      echo '<option value="'.$tier['SSA1Text'].'">'.$tier['SSA1Text'].'</option>'; 
     } 

} 

//************************************** 
//  First selection results  // 
//************************************** 
if($_GET['func'] == "drop_1" && isset($_GET['func'])) { 
    drop_1($_GET['drop_var']); 
} 

function drop_1($drop_var) 
{ 
    include_once('db.php'); 
    $result = mysql_query("SELECT DISTINCT Level FROM qualifications WHERE SSA1Text='$drop_var' ORDER BY Level ASC") 
    or die(mysql_error()); 

    echo '<select name="drop_2" id="drop_2"> 
      <option value=" " disabled="disabled" selected="selected">Choose one</option>'; 

      while($drop_2 = mysql_fetch_array($result)) 
      { 
       echo '<option value="'.$drop_2['Level'].'">'.$drop_2['Level'].'</option>'; 
      } 

    echo '</select>'; 
    echo "<script type=\"text/javascript\"> 
    $('#wait_2').hide(); 
    $('#drop_2').change(function(){ 

     $('#wait_2').show(); 
     $('#result_2').hide(); 
     $.get(\"func.php\", { 
     func: \"drop_2\", 
     drop_var: $('#drop_2').val() 
     }, function(response){ 
     $('#result_2').fadeOut(); 
     setTimeout(\"finishAjax_tier_three('result_2', '\"+escape(response)+\"')\", 400); 
     }); 
     return false; 
    }); 
</script>"; 
} 


{//************************************** 
//  Second selection results  // 
//************************************** 
if($_GET['func'] == "drop_2" && isset($_GET['func'])) { 
    drop_2($_GET['drop_var']); 
} 

function drop_2($drop_var) 
{ 
    include_once('db.php'); 
    $result = mysql_query("SELECT * FROM qualifications WHERE Level='$drop_var' AND SSA1Text='$HELP'") 
    or die(mysql_error()); 

    echo '<select name="drop_3" id="drop_3"> 
      <option value=" " disabled="disabled" selected="selected">Choose one</option>'; 

      while($drop_3 = mysql_fetch_array($result)) 
      { 
       echo '<option value="'.$drop_3['Title'].'">'.$drop_3['Title'].'</option>'; 
      } 

    echo '</select> '; 
    echo '<input type="submit" name="submit" value="Submit" />'; 
} 
?>

任何幫助將不勝感激。

來源

2013-05-30 Flying Ferret

你可以將兩個選擇變量傳遞給drop_2函數嗎？

function drop_2($drop_var1, $drop_var2 == null) { 
    if ($drop_var2 !== null) { 
     $sqlSnipet = " AND SSA1Text='$drop_var2'" 
    } 
    $result = 
     mysql_query(
      "SELECT * 
      FROM qualifications 
      WHERE Level='$drop_var' " . 
      $sqlSnipet 
     ) or die(mysql_error()); 
}

來源

2013-05-30 15:22:27 jimmy

鏈接下拉選項不通過拉入先前的值來查詢

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