in my code i'm showing drop down select field list base on user school choice. for example -
if user choose 'School of Economics' from drop down list, i'm showing in another drop down list just the relevant lanes (based on a mysql query).
to do this i have 5 div's, on for etch school:
<div id='a'>
<span><label>Lane</label></span>
<?php
$sql = "SELECT lane_name FROM lane WHERE `lane_school_id` = 1 ORDER BY `lane_name` ASC";
$result = mysql_query($sql);
echo "<select name='lane_name'>";
while ($row = mysql_fetch_array($result)) {
echo "<option value='" . $row['lane_name'] ."'>" . $row['lane_name'] ."</option>";
}
echo "</select>";
?>
</div>
<div id='b'>
<span><label>Lane</label></span>
<?php
$sql = "SELECT lane_name FROM lane WHERE `lane_school_id` = 2 ORDER BY `lane_name` ASC";
$result = mysql_query($sql);
echo "<select name='lane_name'>";
while ($row = mysql_fetch_array($result)) {
echo "<option value='" . $row['lane_name'] ."'>" . $row['lane_name'] ."</option>";
}
echo "</select>";
?>
</div>
i have a listener to show just the right 'lane' div when user chooses school and hide all other 'lane' div's:
$(document).bind('pageinit', '#indexPage', function(){
$("#a").show();
$("#b").hide();
$("#c").hide();
$("#d").hide();
$("#e").hide();
//this will call our toggleFields function every time the selection value of School field changes
$("#school").change(function () {
toggleFieldsA();
toggleFieldsB();
toggleFieldsC();
toggleFieldsD();
toggleFieldsE();
});
});
function toggleFieldsA() {
if ($("#school").val() == 'School of Economics'){
$("#a").show();
}
else
$("#a").hide();
}
function toggleFieldsB() {
if ($("#school").val() == 'School of Computer Science')
$("#b").show();
else
$("#b").hide();
}
the problem: when user submit the form, i get the wrong 'lane_name' from POST array. i'm getting the last school selected lane value (that is hidden from the user) and not the user selected lane name , plz help
You need to give each <select> element a unique name, otherwise the last element with a particular name will be the only one you can access. PHP has no way of determining which of the 5 lane_name you are referring to.
To figure out what option set was selected you could make the first option blank: <option value=""></option>. You can add some logic to reset to the blank option when you hide/show the DIVs so only one of the lane_names will have any data in it.
The mysql_* is depreciated as of PHP 5.5.0. You should consider switching to mysqli.
Related
I have a form where user selects the category while adding the product.
When user want to edit the product, i am displaying all the previously populated values but could not able to figure out how to display the category he selected.
addproduct.php (displaying the categories from the database)- this code is working fine and can see all the categories in dropdown
<?php
require'dbconn.php';
$subject = mysql_query("select * from categories", $link);;
while($subjectData = mysql_fetch_array($subject)){
echo $subjectData['value'];?>
<option value="<?php echo $subjectData['name'];?>"><?php echo
$subjectData['name'];?>
</option>
In the edit product i want to display all the categories like above, but want to display the selected category in the form which i could not able to do.
editproduct.php (rough draft code) -- not working
<?php
require'dbconn.php';
$subject = mysql_query("select * from categories", $link);;
while($subjectData = mysql_fetch_array($subject)){
echo $subjectData['value'];?>
<option select="<?php echo $cat;?>"value="<?php echo $subjectData['name'];?>"><?php echo
$subjectData['name'];?>
</option>
$cat - category value(previously selected) pulled from database
require'dbconn.php';
$subject = mysql_query("select * from categories", $link);;
<option value="<?php echo $cat;?>"><?php echo $cat;?></option>
while($subjectData = mysql_fetch_array($subject)){
echo $subjectData['value'];?>
<?php if($cat!=$subjectData['name']){?> <option value="<?
php echo $subjectData['name'];?>"><?php echo
$subjectData['name'];?>
</option>
<?php } ?>
Try using this code and please use mysqli as mysql is deprecated. previously selected category should be before while loop. Hope it helps
Two issues with your code:
You are using mysql functions, which are depreciated and don't even exist in the current version of PHP. Use mysqli or PDO functions.
The html you are generating is invalid syntax.
I'll leave the first issue to you to correct.
For the 2nd issue, all of the non-selected options in your dropdown will not have the selected attribute.
Only the selected item will have that attribute. The code below assumes that the variable $cat has the previously selected value, and each row has a
column named 'cat'. When $cat matches the value in the column 'cat', it will add selected='selected' to the option.
<?php
require 'dbconn.php';
$subject = mysql_query("select * from categories", $link);;
while($subjectData = mysql_fetch_array($subject)){
echo $subjectData['value'];
$selected = "";
if($cat == $subjectData['cat']) {
$selected = "selected='selected' ";
}
echo "<option ".$selected."value=".$subjectData['name'].">";
echo $subjectData['name'];
echo "</option>\n";
}
?>
I'm trying to make an invoice php page which contain 2 dropdown option: Supplier_Name and Item_ID. Both of them is not predetermined. Supplier_Name option obtained from another page (List of Supplier, which can be modify). Item_ID option obtained from another page too (List of Item, which can be modify too). I'm quite new about this and my data is small, so I'm looking the simplest way possible.
So far, I knew how to populate Supplier_Name from List of Supplier database. I used this code:
<td>Supplier</td>
<td>
<?php
mysql_connect("localhost","root","");
mysql_select_db("stock");
$result = mysql_query("SELECT * from input_supplier_data");
$jsArray = "var invoice = new Array();\n";
echo '<select name="supplier_name" onchange="changeValue(this.value)">';
echo '<option></option>';
while ($row = mysql_fetch_array($result)) {
echo '<option value="' . $row['supplier_name'] . '">' . $row['supplier_name'] . '</option>';
$jsArray .= "invoice['" . $row['supplier'] . "'] = {name:'" . addslashes($row['supplier']) . "',desc:'".addslashes($row[''])."'};\n";
}
echo '</select>';
?>
</td>
<br /<input type="text" name="input_supplier_data" id="input_supplier_data"/>
<script type="text/javascript">
<?php echo $jsArray; ?>
function changeValue(id){
document.getElementById('input_supplier_data').value = supplier_name[id].name;
};
</script>
When a Supplier selected, only their product will be displayed in the second dropdown option. Since List of Item page contain all item from all supplier, I don't know how to limit them. What I can do so far is only if the supplier was predetermined. I used this code:
SELECT input_item_data.`item_id` FROM `input_item_data` WHERE input_item_data.`supplier` = 'UNILEVER'
But as I stated earlier, the item is not predetermined, it can be modify in List of Item page. Hope anyone can help me. Thanks.
I have two dropdowns in my form.
1)Brand Name of car
2)Model name in which data is going to be retrieved from database on the basis of brand name selected.
so I have taken a div tag and appended that second dropdown.The data in dropdown is displayed properly but the problem is in storing the data selected from the second dropdown, It stores the model name upto the space only.length in the database for that field is also taken properly.
eg- if the model name is "series 1" than it will store only series.
Php page of second dropdown:
<?php
$brand = $_GET['brand'];
include('connection.php');
$sql = "select * from modelname where Brand_Name='" . $brand . "'";
$result = mysqli_query($conn, $sql);
$html = "";
$html = '<select name="model">';
while ($row = mysqli_fetch_assoc($result)) {
$html.='<option value=' . $row['M_Name'] . '>' . $row['M_Name'] . '</option>';
}
$html.='</select>';
echo $html;
?>
because you are not quoting your attributes.
value=hello world
is seen as
value="hello" world
Add the missing quotes.
What I would like to do is have more than one drop down menu in an HTML form and concatenate them so that the choices shown in the second menu are tied to the database element selected in the first menu. This is a snippet of my code:
<div class="form-group">
<?php
$con = mysqli_connect("localhost", "root", NULL) or die("Connection Failed");
mysqli_select_db($con, "tag-it")or die("Connection Failed");
$query = "SELECT * FROM tipologie";
$result = mysqli_query($con, $query); ?>
<select name="tipologia" onchange="">
<?php while ($line = mysqli_fetch_array($result, MYSQL_ASSOC)) { ?>
<option value="<?php echo $line['ID Tipologia'];?>"> <?php echo $line['Nome'];?> </option>
<?php } ?>
</select>
</div>
What it does right now is show all the rows I get from the database in the drop down, but what I want next is to show a second menu that gives me the choices based on the SQL relations and the selected element in the first menu. So I would like to use the option value field (which is a key in my DB) to perform another query and generate a second list. How can this be done? I was thinking of calling a PHP script from the onchange field of the select element, but wouldn't that reload the page?
I am working on a very basic administrator functionality of a social network and I came across this issue of not being able to remove an option from select dropdown list that I previously generated using jquery. The dropdown list contains all users of the social network. Administrator upon clicking on "Delete account" deletes the corresponding record from the database.
Now the question being - when I click on "delete account" it works perfectly fine but the option with a username is still there in a dropdown list and is possible to be picked - when picked it obviously returns dozens of PHP warnings and errors because the record is not in a database anymore. How can I remove this option straight away? I tried something like the following, but it doesn't work.
admin_panel.php (only relevant stuff)
<select name='users' id='users'>
<option value="" disabled selected>Select user</option>
<?php
$sql = mysql_query("SELECT * FROM users WHERE id <>'".$_SESSION['user_id']."'ORDER BY username DESC") or die(mysql_error());
$userList = [];
while($row=mysql_fetch_assoc($sql)){
$username = $row['username'];
$userID = $row['id'];
$userList .= '<option name="userID" value='.$userID.'>'.$username.'</option>';
}
echo $userList;
?>
</select></br></br></div>
<div id="user_info">
<!-- generated user info table-->
</div>
<script type="text/javascript">
"$('#user_info').on('click', '#deleteAccount', function(e){
data.command = 'deleteAccount'
data.userID = $('#users').val()
$.post(theURL, data, function(result){
//Do what you want with the response.
$('#delete_account_success').html(result);
})
$("#users option[value='data.userID']").remove();
$('#delete_account_success').show();
$('#delete_account_success').fadeOut(5000);
})
</script>
processUser.php (part of a switch statement)
if(isset($_POST['command'])){
$cmd = $_POST['command'];
$userID = $_POST['userID'];
$sql=mysql_query("SELECT * FROM users WHERE id='".$userID."'");
$userData = [];
while($row = mysql_fetch_assoc($sql)){
$userData['userid'] = $row['id'];
$userData['username'] = $row['username'];
$userData['name'] = $row['name'];
$userData['date'] = $row['date'];
$userData['email'] = $row['email'];
$userData['avatar'] = $row['avatar'];
$userData['about'] = $row['about'];
$userData['admin'] = $row['admin'];
}
switch($cmd){
case 'deleteAccount':
$sql= "DELETE FROM users WHERE id =".$userID;
$result=mysql_query($sql);
echo "<img src='pics/ok.png' class='admin_updated_ok'>";
break;
}
On this line
$("#users option[value='data.userID']").remove();
You're removing any option items from #users where the value is equal to the string literal data.userID
Try changing it to
$("#users option[value='" + data.userID + "']").remove();