I am trying to edit my form. I want to get selected value selected in selection list.
I have created function to store values in database, and it works. Below is html code I use and function below to insert values in database.
// insert values in database
<label>Dobavljač</label>
<select class="form-control" name="dobavljac" required>
<?php dobavljac() ?>
</select>
function dobavljac(){
$sql=mysqli_query($link, "SELECT * FROM `partneri` WHERE `Dobavljac`='1'
order by `PartnerId` asc ");
echo '<option value="">Izaberi dobavljača</option>';
while($record = mysqli_fetch_array($sql)) {
echo '<option value= "' .$record['PartnerId']. '">' . $record['PartnerNaziv'] . ' </option>';
}
}
// edit values
First I retrieve information from database
$id=$_GET['id'];
$sql = "SELECT * FROM materijali where Id=$id ";
$q = $conn->query($sql);
$r = $q ->fetch();
if ($r) {
$dobavljac=$r['Dobavljac'];
I want to get selected value in box
<label>Dobavljač</label>
<select class="form-control" name="dobavljac" value="<?php echo $dobavljac; ?>">
<?php dobavljac() ?>
</select>
Probably I am not doing it the right way, any advice would be appreciated
Try this
<label>Dobavljač</label>
<select class="form-control" name="dobavljac" value="<?php echo $dobavljac; selected?>">
<option value=<?php echo $dobavljac?> selected>
<?php dobavljac() ?>
</option>
</select>
Try this...
$id=$_GET['id'];
$sql = "SELECT * FROM materijali where Id=$id ";
$q = mysql_query($query);
echo "<select name="dobavljac" class="form-control">";
while (($row = mysql_fetch_row($q)) != null)
{
echo "<option value = '{$row['Dobavljac']}'>";
echo $row['Dobavljac'];
echo "</option>";
}
echo "</select>";
Related
I have made a dropdown search form that is auto populated by my database content. The voices in the table would be for example types of woods with varing dimensions.
So there are repeatable wood names with diverse data.
To avoid repetition the dropdown is populated with wood types combined to be selected then displayed with all their variants.
The problem is, upon selecting an input, the results are of the item listed above and not the one selected.
<form action="search2.php" method="POST">
<select name="finit" onchange='this.form.submit()'>
<?php
include("connect.php");
$query = "SELECT finit FROM prime";
$info = mysqli_query($conn, $query);
$finit = '';
echo "<option value=\"\">Selezione Materiale</option>";
while($row = $info->fetch_assoc()){
if($row['finit'] != $finit) {
echo "<option value=\"$finit\">" . $row['finit'] . "</option>";
$finit = $row['finit'];
}
}
?>
</select>
<noscript><input type="submit" value="Submit"></noscript>
</form>
Since there are many variants(dimensions) associated with a single wood type, you have to first take the wood type as input from user(via dropdown list), and probably then you may want to display all possible variants(dimensions) of that particular wood type.
So, change the SQL query in the following way,
$query = "SELECT DISTINCT finit FROM prime";
and the while loop in the following way,
while($row = $info->fetch_assoc()){
$output = "<option value='" . $row['finit'] . "'";
if($row['finit'] == $_POST['finit']){
$output .= " selected='selected'";
}
$output .= ">" . $row['finit'] . "</option>";
echo $output;
}
Try this but change if condition according your default value and sql value should be match.
<form action="search2.php" method="POST">
<select name="finit" onchange='this.form.submit()'>
<?php
include("connect.php");
$query = "SELECT finit FROM prime";
$info = mysqli_query($conn, $query);
$finit = '';
?>
<option value="">Selezione Materiale</option>;
<?php
while($row = $info->fetch_assoc()){
if($row['finit'] == $finit) {
$selected = 'selected';
}else{
$selected = '';
$finit = $row['finit'];
}
?>
<option value="<?php echo $finit ?>" <?php echo $selected ?>><?php echo $row['finit']?></option>
<?php } ?>
</select>
<noscript><input type="submit" value="Submit"></noscript>
</form>
how to fix this syntax for this logic ?
i want to select my select option to select the another select option
<?php
$query_string = "SELECT * FROM products";
$query_string1 = "SELECT * FROM suppliers where ProductID = // firstSelectoption(value)";
$query_string2 = "SELECT * FROM categories";
$query = mysql_query($query_string);
$query1 = mysql_query($query_string1);
$query2 = mysql_query($query_string2);
?>
and in the body i make
<select name="first" id="first" onchange="childrenOnChange(this.value)">
<?php
while ($row = mysql_fetch_array($query)) {
echo '<option value=' . $row["ProductID"] . '>';
echo $row['ProductID'];
echo '</option>';
}
?>
</select>
<select name="second" id="second">
<?php
while ($row = mysql_fetch_array($query1)) {
echo '<script>';
echo 'var arr = array(';
$row['SupplierID'] . ',';
echo ')';
echo '</script>';
}
?>
</select>
i want to set the second select option value with $query1;
If you get the value from the query with $val = mysql_fetch_array($query1), then you can include the following in your loop:
$selected = '';
if ($row['ProductID'] == $val) {
$selected = "selected";
}
echo '<option value="'.$row['ProductID'].'" '.$selected.'>'.$row['ProductID'].'</option>';
I have select option. this option has multiple value from database. I want to update something from database, this value i want to update is exist on the select option I have.
this is my option code
$id = $_GET['update'];
$query = mysql_query("SELECT * FROM transaction where id = '$id'") or die ("could not search");
$count = mysql_num_rows($query);
while ($rows = mysql_fetch_array($query)) {
$id = $rows['id'];
$tranid = $rows['tranid'];
$trandate = $rows['trandate'];
$patientid = $rows['patientid'];
$transactiontype = $rows['transactiontype'];
$trandescription = $rows['trandescription'];
$tranquantity = $rows['tranquantity'];
$tranunitprice = $rows['tranunitprice'];
$tranamount =$rows['tranamount'];
$gettrandescription = $rows['trandescription'];
}
}
if (isset($_POST['selectmedicine'])) {
$gettrandescription=$_POST['medicineid'];
}
if (isset($_POST['selectroomquantity'])) {
$tranquantity=$_POST['quantity'];
}
?>
<script type="text/javascript">
$('#collapseone').collapseone({
toggle: true
});
<option value="<?php echo $trandescription; ?>" <?php if($trandescription==$gettrandescription){ echo "selected";} ?> ><?php echo $gettrandescription; ?></option>
<option value="<?php echo $tranquantity; ?>" <?php if($tranquantity==$tranquantity){ echo "selected";} ?> ><?php echo $tranquantity; ?></option>
this has value results, but i cant fetch this value to my existing select option.
If you want to "make that variable an array" as aldrin27 said, append [] to the name attribute of the select tag. The selected value of the option with name selectroomquantity will be available in your script as $_POST["selectroomquantity"] (this is the varible).
<select multiple name="selectroomquantity[]">
<option value="...">...</option>
</select>
It should only be necessary if multiple options can be selected however.
Also, there seems to be a typo:
<?php if($tranquantity==$tranquantity)
That check will always return true. It should probably be:
<?php if($tranquantity==$gettranquantity)
hi all i just got the code on how to fecth the value to dropdown. actually i made a wrong word. pre-selected is the right one, sorry for that. here;s the working code.
<select name="selectmedicine" class="form-control col-sm-4" id="medicinename">
<option id="0" style="width:100px"></option>
<?php
$medicine = mysql_query("SELECT * FROM medicine");
while ($row = mysql_fetch_array($medicine)) {
echo '<option id="' . $row['medicinename'] . '"';
echo ' value="' . $row['medicineid'] . '"';
if($row['medicinename'] == $trandescription) {
echo ' selected="selected"';
}
echo '>';
echo $row['medicinename'];
echo '</option>';
}
?>
</select>
thanks everyone, whos trying to help me on this. actually this is my five revised question sorry for that. and finally i got the right one.
I'm trying to do a select from a table based on the post value of an HTML select box. I'm getting no results at all, I'm echoing out the post value no problem. The statement works on it's own but won't when I use the select form to populate it. This is just my test I will be adding other options to the dropdown box.
<?php
if(isset($_POST['value'])) {
if($_POST['value'] == 'Militaria') {
$query = "SELECT * FROM listings WHERE category1=Militaria";
}
else {
// query to get all records
$query = "SELECT * FROM listings";
}
}
$sql = mysql_query($query);
while ($row = mysql_fetch_array($query)){
echo 'Description:' . $row['description'];
}
mysql_close($con);
?>
Here is the html form I'm using, can anyone tell me where I'm going wrong, should I do it a different way etc, I'm new to php? Thanks!!
<form action='<?php echo $_SERVER['PHP_SELF']; ?>' method='post' name='form_filter' >
<select name="value">
<option value="all">All</option>
<option value="Militaria">Militaria</option>
</select>
<br />
<input type='submit' value = 'Filter'>
</form>
mysql_fetch_array() should receive resorce as a parameter. Try mysql_fetch_array($sql).
Quote around 'Militaria' and mysql_fetch_array($sql)
<?php
if(isset($_POST['value'])) {
if($_POST['value'] == 'Militaria') {
$query = "SELECT * FROM listings WHERE category1='Militaria'";
}
else {
// query to get all records
$query = "SELECT * FROM listings";
}
$sql = mysql_query($sql);
while ($row = mysql_fetch_array($sql)){
echo 'Description:' . $row['description'];
}
mysql_close($con);
}
?>
<form action='<?php echo $_SERVER['PHP_SELF']; ?>' method='post' name='form_filter' >
<select name="value">
<option value="all">All</option>
<option value="Militaria">Militaria</option>
</select>
<br />
<input type='submit' value = 'Filter'>
</form>
You have two mistakes in your php code.
1st : quote around Militaria. The query should be, $query = "SELECT * FROM listings WHERE category1='Militaria'";
2nd : mysql_fetch_array accepts executed query's result as parameter. It should be, $row = mysql_fetch_array($sql)
Final code:
<?php
if(isset($_POST['value'])) {
if($_POST['value'] == 'Militaria') {
$query = "SELECT * FROM listings WHERE category1 = 'Militaria'";
}
else {
// query to get all records
$query = "SELECT * FROM listings";
}
}
$sql = mysql_query($query);
while ($row = mysql_fetch_array($sql)){
echo 'Description:' . $row['description'];
}
mysql_close($con);
?>
We managed to get the drop down list menu however, we are having difficulties getting the data from sql. So far, this is what we got.
<select>
<option id="">--Select jobscope--</option>
<?php
$con=mysqli_connect("host","user","password","database");
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$getIT = mysqli_query("SELECT job_title FROM `job_details`");
while($viewIT = mysqli_fetch_array($getIT)) {
}
?>
<option id="<?php echo $viewIT['job_title']?>"<?php echo $viewIT['job_title']?></option>
</select>
Shouldn't be like this ? with tag inside WHILE LOOP
while($viewIT = mysql_fetch_array($getIT)) {
<option id="<?php echo $viewIT['job_title']?>"<?php echo $viewIT['job_title']?></option>
}
$query = "SELECT * FROM test_groups_tb WHERE user_id='$userid'";
$result = mysql_query($query) or die(mysql_error());
while($row = mysql_fetch_assoc($result))
{
$dd .= "<option value='{$row['group_id']}'>{$row['group_name']}</option>";
}
Try this,
<option value="">--select--</option>
<?php
while($rec = mysql_fetch_assoc($result)) {
?>
<option value="<?=$rec['job_title']?>"><?=$rec['job_title']?></option>
<?php }
}?>
</select>
I am not from php background. Try this.
<?php
$query = "SELECT job_title FROM job_details";
$result = $mysqli->query( $query );
echo '<select id="domain_account" name="domain_account" class="txtBox">';
echo '<option value="">-select-</option>';
while ($row = $result->fetch_assoc()){
?>
<option value="<?php echo $row['job_title']; ?>"><?php echo $row['job_title']; ?></option>
<?php
}
echo "</select>";
?>
Better use PDO or MYSQLi . MYSQL* is depriciated
U need to use that <option id="<?php echo $viewIT['job_title']?>"<?php echo $viewIT['job_title']?></option> line b/w while loop
$getIT = mysql_query("SELECT job_title FROM `job_details`");
while($viewIT = mysql_fetch_array($getIT)) {?>
<option id="<?php echo $viewIT['job_title']?>"<?php echo $viewIT['job_title']?></option>
<?hp }?>