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I have no knowledge of javascript, and would like to write a function that queries my database using onchange. I have a dropdown which shows my mysql tables. when selected (and pushing the select button) it queries the database, and shows the contents of the table. How do I get rid of that submit button, and start using onchange. I tried searching around, but apparently this is a difficult topic to find. Anyone have any ideas?
hi,
i'm presuming you simply wish to send the tablename as a variable to another page. if you wish to include it within a sql statement, you only need the second parameter (tblName)
VinnyCode:<form name='theform'> <select name='thesel' onchange='sendit(this.form, this.options[this.selectedIndex].value)'> <option value='someTableName'>someTableName</option> ....more options.... </select> </form> function sendit(formObj, tblName) { formObj.action = 'somepage.php?table=' + tblName; formObj.submit(); }
moderator at:Webxpertz Forums
let me try to be more descriptive....I currently have a dropdown, which queries, and shows a list of my tables. Upon selection (submit button) appears everything in the table under the dropdown. everything's all in one page with some if statements. How can I use the onchange to get it to do the same thing, without needing the submit button? Check out the link to view the page I'm talking about. I didnt really understand your reply, so I can't tell if its what I'm asking for or not.
http://dky.sytes.net/golesnew/cat/showtable.php
I'm no expert, but Vinny's answer looks to be exactly what you're asking for: using the onchange event on the <select> to call a function which constructs the action attribute and fires the form submit.
gav
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I think you could just add this to the <select>:Code:onchange="this.form.submit();"
Who walks the stairs without a care
It shoots so high in the sky.
Bounce up and down just like a clown.
Everyone knows its Slinky.
but what takes the place of the this, and what takes the place of the form and the submit.... I want it to be submitted to the current form.
Nothing takes their place. They are used literally to indicate the current form.
Of course, that's just my opinion. I could be wrong.
this is a magic word
</spam>
Think of 'this' as a pointer. It points to whatever you are pointing to. So,
<form name='a'>
<select name='b' onchange=doit(this, this.form, this.selectedIndex)
the 1st 'this' means the selection list object
the 2nd 'this' means form 'a'
the 3rd 'this' means the index of the selected option in select 'b'
Vinny
moderator at:Webxpertz Forums
This makes perfect sense now, onchange points to the action in the form, I got it. and if I want, I can put it like so mypage.php?mode=change in the action, and then use a php if statement to run the query, like so.That's awesome!!! Thanks a lot everyone. One problem I have though, my dropdown disappears when selected. here's the page I'm working on. http://dky.sytes.net/golesnew/cat/pickitem.php and here's the code, in case anyone knows a little php, and wants to see if they can tell my where my dropdown went??PHP Code:if('change' == $_GET['mode'])
{
}
PHP Code:<html><head><title>Pick an Item Page</title><LINK HREF="catalogwhite.css" REL="STYLESHEET"></head><body>
<?
require("connect.php" );
database_connect();
display_tables();
if (!$_GET)
{
echo "<form action=\"$_SERVER['PHP_SELF']?mode=change\" method=\"post\" NAME=\"show_tables\">";
echo "<br />";
echo "<table align=\"center\">";
echo "<tr>";
echo "<td width=\"100%\" valign=\"top\" align=\"center\">";
# must add this in the select below onChange=\"location.href="?mode=change&table=" . $table . ''\"
echo "<select name=\"view_category\" onchange=\"this.form.submit();\">";
echo "<option value=\"ViewTheContentsofaTable\"> ";
echo " View a Category ";
echo " ";
echo "</option>;";
while($row= mysql_fetch_array($result)) {
echo "\t\t\t\t<option>{$row[0]}</option>\n";
}
echo "</select>";
#echo " <input type=\"submit\" name=\"Select\" value=\"Select\" />";
echo "</td></tr></table>";
}
#if ($_POST["Select"]) {
#$table = $_POST["view_category"];
if('change' == $_GET['mode']) {
$table = $_POST["view_category"];
# $table=$_GET['table'];
if ($table == "ViewTheContentsofaTable" )
{
echo "<br><br><br><br><font id='title'>Please Select a Category</font>";
}
else
{
$query = "SELECT * FROM $table";
$result = mysql_query($query);
if (!$result)
{
$mcrmessage = mysql_error();
echo "$mcrmessage<br>";
die();
}
ECHO "<TABLE BORDERCOLOR='#003300' BORDER='0' CELLSPACING='0' CELLPADDING='10'>";
echo "<tr><td colspan='2' STYLE='border-bottom-style: solid; border-bottom-width:1'><center>
<font id='title'>$table</font></td></tr>";
while ($row = mysql_fetch_array($result))
{
extract($row);
echo "<tr onclick=\"java script:top.showitem.location.href('showitem.php?mode=view&
table=".$table."&main_id=".$row["Product_Number"]."');\" STYLE=\"Cursor:Hand;\"
ONMOUSEOVER=\"this.style.backgroundColor='#E2EEE2';\"
ONMOUSEOUT=\"this.style.backgroundColor='#FFFFFF';\">
<td STYLE='border-bottom-style: solid; border-bottom-width:1'>$Thumbnail</td>
<td STYLE='border-bottom-style: solid; border-bottom-width:1'>$Item_Name</td></tr>\n";
}
echo "</table>\n";
}
}
database_disconnect();
?>
</form></body></html>
I tried your page on Mozilla 1.5 and IE6 and it seems to work just fine.
Of course, that's just my opinion. I could be wrong.
yeah, it works fine, but like i said the dropdown disappears when you select something, any ideas why?
Well, the dropdown doesn’t exactly “disappear.” Selecting something in the dropdown submits the form, which in turn takes you to the form-submission page. Since the dropdown isn’t on the form-submission page, it would give the effect of it “disappearing.”
Of course, that's just my opinion. I could be wrong.
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