[Linux-programlama] Cross Select Box Kod örneği

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From: Erdal YAZICIOGLU (erdal.yazicioglu@gmail.com)
Date: Wed 29 Dec 2004 - 11:36:18 EET


Merhabalar,

belki bazı arkadaşların işine yarabilir düşünce ile iki farklı tablodan iki select box seçimi yapmaya yarayan bir kod gönderiyorum. Birinci seçim sonrası ikinci select box otomatik olarak değişmekte...Kaynak:http://www.phpaddict.com/forum2/400.html

<?PHP
 $Host = "127.0.0.1";
 $User = "root";
 $Passwd = "";
 $DBName = "caucasia";
 $TableName = "dict_hospitals";

 $Link = mysql_connect($Host, $User, $Passwd) or die("Could not connect: " . mysql_error());

 $Query = "SELECT hospitalID,hospitalName ".
    "FROM $TableName";
 $Result = mysql_db_query( $DBName, $Query, $Link );
?>
<HTML>
<HEAD>
 <SCRIPT language="JavaScript">
 <!--
  function BodyLoad() {
   var select = document.FormName.Main_category;
   select.options[0] = new Option("Choose One");
   select.options[0].value = 0;
   <?PHP
    $ctr = 1;
    While( $Row = mysql_fetch_array($Result) ) {
     echo "select.options[$ctr] = new Option(\"$Row[hospitalName]\");\n";
     echo "select.options[$ctr].value = \"$Row[hospitalID]\";\n";
     $ctr++;
    }
   ?>
  }

  function Fill_Sub() {
   var main_select = document.FormName.Main_category;
   var sub_select = document.FormName.Sub_category;
   if( main_select.options[main_select.selectedIndex].value != 0 ) {
    sub_select.length = 0;
   }
   <?PHP
    $Query = "SELECT hospitalID,hospitalName ".
       "FROM $TableName";
    $Result = mysql_db_query( $DBName, $Query, $Link );

    while( $Row = mysql_fetch_array($Result) ) {
   ?>
   if( main_select.options[main_select.selectedIndex].text == "<?PHP echo $Row[hospitalName]; ?>" ) {
    <?PHP
     $Query2 = "SELECT departmentID,departmentName ".
        "FROM dict_hospdepartments ".
        "WHERE hospitalID = '$Row[hospitalID]'";
     $Result2 = mysql_db_query( $DBName, $Query2, $Link );

     $ctr = 0;
     While( $Row2 = mysql_fetch_array($Result2) ) {
      echo "sub_select.options[$ctr] = new Option(\"$Row2[departmentName]\");\n";
      echo "sub_select.options[$ctr].value = \"$Row2[departmentID]\";\n";
      $ctr++;
     }
    ?>
   }

   <?PHP
    }

    mysql_close($Link);
   ?>
  }
 -->
 </SCRIPT>
</HEAD>
<BODY onload="BodyLoad();">
 <FORM name="FormName" method="POST" action="">
  <TABLE border="1">
   <TR>
    <TD>Main Category</TD>
    <TD>Sub Category</TD>
   </TR>
   <TR>
    <TD>
     <SELECT name="Main_category" onchange="Fill_Sub();"></SELECT>
    </TD>

    <TD>
     <SELECT name="Sub_category" size="4"></SELECT>
    </TD>
   </TR>
  </TABLE>
 </FORM>
</BODY>
</HTML>

Erdal YAZICIOĞLU

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