hi friends,
i want to display data from mySQL database, and iam using two drop down list, displaying data in second drop down list depending on the selecting from first drop down list, for example:
first drop down list contains states, and second drop down list contains cities.
almualim 0 Light Poster
nav33n 472 Purple hazed! Team Colleague Featured Poster
This has been discussed so many times. Either use ajax or this way. In the first dropdown list, list all the states. Have an onchange event to submit the form for the 1st dropdown list. When a user selects an option from the 1st dropdown list, onchange event will be fired and the form will be submitted. You can then access the selected value of 1st dropdown list by doing $firstdropdownlistvalue = $_POST['dropdownlistname']; You can then query the database for this particular state and get all the cities in the 2nd dropdown list.
almualim 0 Light Poster
i tried that, but the cities didn't appeared in the second dropdown list.
note:
" the code in the same page ".
nav33n 472 Purple hazed! Team Colleague Featured Poster
Can you post your code here ?
almualim 0 Light Poster
ok, that is:
<html >
<head>
<title> my page </title>
</head>
<body >
<?php
mysql_connect ("localhost","root","");
mysql_select_db ("University");
$get_list_result = mysql_query("select Col_Name from College");
echo '<form name="delete" method="POST" action="delete_dept2.php">';
echo'<select name="col_name">';
while ($recs = mysql_fetch_array($get_list_result)){
$display_list = $recs['Col_Name'];
echo '<option>'; echo ($display_list); echo '</option>';
}
echo '</select>';
$firstdropdownlistvalue = $_POST['col_name'];
$get_list_result2 = mysql_query("select Col_No from College where Col_Name='$firstdropdownlistvalue'");
$get_list_result3= mysql_query("select D_Name from Department where Col_No='$get_list_result2'");
echo'<select name="dept_name">';
while ($recs2 = mysql_fetch_array($get_list_result3)){
$display_list2 = $recs2['D_Name'];
echo '<option>'; echo ($display_list2); echo '</option>';
}
echo '</select>';
?>
</form>
</body>
</html> nav33n 472 Purple hazed! Team Colleague Featured Poster
First of all, you don't have an onchange event to submit the page. secondly, you should have a value for the option. Here is a simple example. I haven't tested it, but it should work with little tuning (which you have to figure it yourself !)
<?php
$con = mysql_connect("localhost","root");
mysql_select_db("test");
$query = "select col1 from table";
$result = mysql_query($query);
echo "<form method='post' action='test.php'>";
echo "<select name=dropdown1 onchange='javascript: document.form.submit();'>";
while($row = mysql_fetch_array($result)) {
$value = $row['col1'];
echo "<option value='".$value."'>$value</option>";
}
echo "</select>";
$firstdropdownlistvalue = $_POST['dropdown1'];
$query2 = "select * from table where col2 = '$firstdropdownlistvalue'";
$result2 = mysql_query($query2);
echo "<select name=dropdown2>";
while($row2 = mysql_fetch_array($result2)) {
$value2 = $row2['col1'];
echo "<option value='".$value2."'>$value2</option>";
}
echo "</select>";
echo "</form>";
?> almualim 0 Light Poster
i tried that but i get one error in this line:
echo "<select name=dropdown1 onchange='javascript: document.form.submit();'>";
the error is:
parse error, unexpected T_STRING, expecting ',' or ';'
nav33n 472 Purple hazed! Team Colleague Featured Poster
Okay ! Well, I just re-wrote the whole code and it works ! Change the column names and tablename to suit your needs ! :) If this doesn't work, then, I don't know, something must be wrong from your side.
<?php
$con = mysql_connect("localhost","root");
mysql_select_db("test");
$firstdropdownlistvalue = $_POST['dropdown1'];
$query = "select order_id from orders";
$result = mysql_query($query);
echo "<form method='post' name='form1' action='test.php'>";
echo "<select name=dropdown1 onchange='javascript: document.form1.submit();'>";
while($row = mysql_fetch_array($result)) {
$value = $row['order_id'];
if($value == $firstdropdownlistvalue) { $selected = "selected"; } else { $selected = ""; }
echo "<option value='".$value."' $selected>".$value."</option>";
}
echo "</select>";
$query2 = "select * from orders where order_id = '$firstdropdownlistvalue'";
$result2 = mysql_query($query2);
echo "<select name=dropdown2>";
while($row2 = mysql_fetch_array($result2)) {
$value2 = $row2['amount'];
echo "<option value='".$value2."'>$value2</option>";
}
echo "</select>";
echo "</form>";
?> almualim 0 Light Poster
thank you very much, the error is solved but still data dosn't appear in the second dropdown list, I appreciate your Effort, please if you know the solution for this problem tell me, and iam Wait you.
thanks...
nav33n 472 Purple hazed! Team Colleague Featured Poster
Hi, the above code works fine. You just have to change the tablenames/column names. Anyway, When the page gets submitted, echo the value of 1st dropdownlist. See if the value is passed correctly. If yes, then the problem must be with your 2nd query. Check your query. See if your query is right. If it is correct, then, something is going wrong while fetching the records from the table. :) You should check this yourself because I dont have the access to your table and I don't know what exactly is the problem.
almualim 0 Light Poster
hi, ok i tried to print this value: $firstdropdownlistvalue = $_POST;
by echo($firstdropdownlistvalue);
but it's doesn't appeared in the browser. why?
nav33n 472 Purple hazed! Team Colleague Featured Poster
Post your latest code.
almualim 0 Light Poster
okay:
<?php
mysql_connect ("localhost","root","");
mysql_select_db ("University");
$firstdropdownlistvalue = $_POST['col_name'];
$get_list_result = mysql_query("select Col_Name from College");
echo '<form name="delete" method="POST" action="delete_dept2.php">';
echo'<select name=col_name onchange=javascript: document.delete.submit(); style="position:absolute;left:311px;top:146px;width:303px;z-index:2">';
while ($recs = mysql_fetch_array($get_list_result)){
$display_list = $recs['Col_Name'];
if($display_list==$firstdropdownlistvalue){
$selected = "selected";
}
else
{ $selected = "";
}
echo "<option value='".$display_list."' $selected>" .$display_list."</option>";
}
echo "</select>";
echo ($firstdropdownlistvalue);
$get_list_result2 = mysql_query("select * from College where Col_Name='$firstdropdownlistvalue'");
$get_list_result3= mysql_query("select D_Name from Department where Col_No='$get_list_result2'");
echo'<select name="formselect2" style="position:absolute;left:310px;top:223px;width:308px;z-index:4">';
while ($recs2 = mysql_fetch_array($get_list_result3)){
$display_list2 = $recs2['D_Name'];
echo "<option value='".$display_list2."'> $display_list2 </option>";
}
echo '</select>';
?> nav33n 472 Purple hazed! Team Colleague Featured Poster
onchange="javascript: document.delete.submit();" instead of
onchange=javascript: document.delete.submit();
Secondly,
$get_list_result2 = mysql_query("select * from College where Col_Name='$firstdropdownlistvalue'");
$get_list_result3= mysql_query("select D_Name from Department where Col_No='$get_list_result2'");
is wrong. What are you trying to do ?
almualim 0 Light Poster
see the last update on the code:
<?php
mysql_connect ("localhost","root","");
mysql_select_db ("University");
$firstdropdownlistvalue = $_POST['col_name'];
$get_list_result = mysql_query("select Col_Name from College");
echo '<form name="delete" method="POST" action="delete_dept2.php">';
echo'<select name=col_name onchange="javascript: document.delete.submit();" style="position:absolute;left:311px;top:146px;width:303px;z-index:2">';
while ($recs = mysql_fetch_array($get_list_result)){
$display_list = $recs['Col_Name'];
if($display_list==$firstdropdownlistvalue){
$selected = "selected";
}
else
{ $selected = "";
}
echo "<option value='".$display_list."' $selected>" .$display_list."</option>";
}
echo "</select>";
echo ($firstdropdownlistvalue);
$get_list_result2 = mysql_query("select col_No from College where Col_Name='$firstdropdownlistvalue'");
while ($recs2 = mysql_fetch_array($get_list_result2)){
$display_list2 = $recs2['col_No'];
}
$get_list_result3= mysql_query("select D_Name from Department where Col_No='$$display_list2'");
echo'<select name="formselect2" style="position:absolute;left:310px;top:223px;width:308px;z-index:4">';
while ($recs3 = mysql_fetch_array($get_list_result3)){
$display_list3 = $recs3['D_Name'];
echo "<option value='".$display_list3."'> $display_list3 </option>";
}
echo '</select>';
?>$get_list_result2 = mysql_query("select col_No from College where Col_Name='$firstdropdownlistvalue'");here i want to get college_no(col_No) from college table, if the college name(col_Name) equals the value came from 1st dropdown list($firstdropdownlistvalue) $get_list_result3= mysql_query("select D_Name from Department where col_No='$display_list2'"); and here i want to get department name(D_Name) from Department table, if college_no(col_No) equals $display_list2
nav33n 472 Purple hazed! Team Colleague Featured Poster
Then use a sub query.
$query = "select D_Name from Department where col_No = (select col_No from College where col_name = '$firstdropdownlistvalue')";This will return the D_name of the selected col_name.
almualim 0 Light Poster
Tired with me, but the problem still exist and I do not know what to do...
nav33n 472 Purple hazed! Team Colleague Featured Poster
Use firefox to execute your scripts. Make use of error console to see any javascript error.
almualim 0 Light Poster
i tried with firefox, but when i was select value from 1st dropdown list Transmission occurs for the second page, but i want the values appear in the same page.
nav33n 472 Purple hazed! Team Colleague Featured Poster
What second page ? Why dont you keep everything in the same page ? :-/
almualim 0 Light Poster
i mean two dropdown lists in same page
nav33n 472 Purple hazed! Team Colleague Featured Poster
Yep.. My example works with 2 dropdowns in a same page.
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