I have an error here, the data retrived from mysql database and listed in listbox or dropdown, when i need to update my form i get the result after refresh.
I have tried but need to sortout
<?
include("connect.php");
mysql_select_db("joblisting") or die(mysql_error());
$result = mysql_query("SELECT * FROM positiontitle") or die(mysql_error());
$ans=$positiontitle;
echo $ans;
?><select size=1 name="positiontitle" id="positiontitle"><?
echo '<option value="'.$ans.'">';
echo "</option>";
while ($row = mysql_fetch_array($result))
{
echo "<option>";
echo $row['positiontitle'];
echo "</option>";
}
echo "<br />";
?>
where are you defining $positiontitle ? shouldn't it be $_POST['positiontitle']?
Related
I have below code which creates drop-down from php. I want to achieve 2 things here.
1. I want to set one of the option to be set default. It may be hard coded or selected from query.
2. When hit button, it should retain selected option. I could retrieve selected option using session data with this: echo $print_version1[array_keys($print_version1)[0]];
Drop-down code:
$result = $conn->query("SELECT DISTINCT nx_version FROM workflow1 ORDER BY id DESC");
echo "<form action='http://localhost/w_5aug/process.php' method='get'>";
echo "<html>";
echo "<body>";
echo "<p></p>";
echo "<center>";
echo "<strong> Select Base Verison To Compare With : </strong>";
echo "<select name='nx_version' id='nx_version'>";
while ($row = $result->fetch_assoc()) {
$nx_version = $row['nx_version'];
echo '<option>'.$nx_version.'</option>';
}
echo "</select>";
echo " <button type='submit'><b>Add Base Verison</b></button>";
echo "</center>";
echo "</body>";
echo "</html>";
echo "<p></p>";
$array_select = $_SESSION['data'];
print_r($array_select);
echo "<form>";
i assume that the option that should be selected is$print_version1[array_keys($print_version1)[0]
while ($row = $result->fetch_assoc()) {
$nx_version = $row['nx_version'];
if($_SESSION["id"]) {
if($nx_version == "the hardcode value you want to be selected"){
echo '<option selected="selected">'.$nx_version.'</option>';
}else{
echo '<option>'.$nx_version.'</option>';
}
}else{
if($print_version1[array_keys($print_version1)[0]] == $nx_version){
echo '<option selected="selected">'.$nx_version.'</option>';
}else{
echo '<option>'.$nx_version.'</option>';
}
}
}
I have below code which creates drop-down from php. I want to achieve 2 things here.
1. I want to set one of the option to be set default. It may be hard coded or selected from query.
2. When hit button, it should retain selected option. I could retrieve selected option using session data with this: echo $print_version1[array_keys($print_version1)[0]];
Drop-down code:
$result = $conn->query("SELECT DISTINCT nx_version FROM workflow1 ORDER BY id DESC");
echo "<form action='http://localhost/w_5aug/process.php' method='get'>";
echo "<html>";
echo "<body>";
echo "<p></p>";
echo "<center>";
echo "<strong> Select Base Verison To Compare With : </strong>";
echo "<select name='nx_version' id='nx_version'>";
while ($row = $result->fetch_assoc()) {
$nx_version = $row['nx_version'];
echo '<option>'.$nx_version.'</option>';
}
echo "</select>";
echo " <button type='submit'><b>Add Base Verison</b></button>";
echo "</center>";
echo "</body>";
echo "</html>";
echo "<p></p>";
$array_select = $_SESSION['data'];
print_r($array_select);
echo "<form>";
i assume that the option that should be selected is$print_version1[array_keys($print_version1)[0]
while ($row = $result->fetch_assoc()) {
$nx_version = $row['nx_version'];
if($_SESSION["id"]) {
if($nx_version == "the hardcode value you want to be selected"){
echo '<option selected="selected">'.$nx_version.'</option>';
}else{
echo '<option>'.$nx_version.'</option>';
}
}else{
if($print_version1[array_keys($print_version1)[0]] == $nx_version){
echo '<option selected="selected">'.$nx_version.'</option>';
}else{
echo '<option>'.$nx_version.'</option>';
}
}
}
<?php
$id = $_SESSION['user_id'] ;
echo "<form method='post' action='#'>";
echo "</select>
<p>Which Hospital Would You Like to Submit To?</p>";
$queryitem = "SELECT * FROM vendor_hospital WHERE vendor_hospital.user_id = '$id' AND vendor_hospital.approval_status = '1'" or die('MYSQL error: ' . mysql_error());
if ($result = mysql_query($queryitem)) {
if ($success = mysql_num_rows($result) > 0) {
echo "<select name='hospital_name'>";
echo "<option>-- Select A Facility --</option>";
while ($row = mysql_fetch_array($result))
echo "<option value='$row[manufacturer_id]'>$row[manufacturer_id]</option>";
echo "</select><br><br>";
} else {
echo "No results found.";
}
} else {
echo "Failed to connect to database.";
}
echo "<input type='submit' value='Submit' name='submit' class='button' /></form>";
?>
For some reason I'm stuck here. I'm just trying to get the manufacturer name to show in my options instead of the manufacturer_id. The manufacturer name is a foreign key in another table so I can't simply call $row[manufacturer_id] in my option tag. What should I do here? My only thought is to run a query inside the option tag for every manufacturer_id listed as a value but I'm sure that is overkill. Can someone point me in the right direction of a more elegant solution than that?
Not really sure what you are trying to do...but try this.
Replace this:
while ($row = mysql_fetch_array($result))
echo "<option value='$row[manufacturer_id]'>$row[manufacturer_id]</option>";
echo "</select><br><br>";
With this:
while ($row = mysql_fetch_array($result)){
echo "<option value='".$row['manufacturer_id']."'>".$row['manufacturer_id']."</option>";
}
echo "</select><br><br>";
Also be sure to double check and make sure that you are pulling from the right database, that it is populated, you are calling the right table names, etc...
I am new to php, i created drop down which calling data from mysql data base, user selects option and its save to data base.
Problem Arises in edit form in which its do not showing selected value.
Drop Down code is below:
$query = 'SELECT name FROM owner';
$result = mysql_query($query) or die ('Error in query: $query. ' . mysql_error());
//create selection list
echo "<select name='owner'>\name";
while($row = mysql_fetch_row($result))
{
$heading = $row[0];
echo "<option value='$heading'>$heading\n";
}
echo "</select>"
Please advise solution for the edit form.
Thanks in Advance
you must close <option> tag:
echo "<option value='$heading'>$heading</option>";
$query = 'SELECT name FROM owner';
$result = mysql_query($query) or die ('Error in query: $query. ' . mysql_error());
//create selection list
echo "<select name='owner'>\name";
while($row = mysql_fetch_row($result))
{
$heading = $row[0];
?>
<option <?php if($heading=="SOMETHING") { echo "selected='selected'"; } ?> value="SOMETHING">SOMETHING</option>
<option <?php if($heading=="SOMETHING2") { echo "selected='selected'"; } ?> value="SOMETHING2">SOMETHING2</option>
<option <?php if($heading=="SOMETHING3") { echo "selected='selected'"; } ?> value="SOMETHING3">SOMETHING3</option>
<?php
}
echo "</select>"
I'd do it this way.
$numrows = mysql_num_rows($result);
if ($numrows != 0){
echo "<select name='owner'>\name";
while ($x = mysql_fetch_assoc($result)){
echo "<option value='".$x['heading']."'>".$x['heading']."</option>";
}
echo "</select>";
}
$x['heading'] is using the value of the row 'heading' in the database
It's much more efficient and simply looks more sophisticated.
What i'm trying to do is display a drop down with all field names from mysql database, once the user picks one and submits the form i want to display a second dropdown filled with all the rows from the submitted field name, this is my code so far:
$result = mysql_query("select * from `parts`") or die(mysql_error());
echo "<form action='".$_SERVER['PHP_SELF']."' method='post'>";
echo "<select name='field_names'>";
$i = 0;
while ($i < mysql_num_fields($result)) {
$fieldname = mysql_field_name($result, $i);
echo '<option value="'.$fieldname.'">'.$fieldname.'</option>';
$i++;
}
echo "</select>";
echo "<input type='submit' value='submit'></input>";
echo "</form>";
if($_POST) {
$fields = $_POST['field_names'];
$result1 = mysql_query("select '".$fields."' from `parts`") or die(mysql_error());
echo '<select name="fields">';
while ($row = mysql_fetch_array($result1)) {
echo "<option value=".$row[$fields].">".$row[$fields]."</option>";
}
echo '</select>';
}
Can anyone spot where i'm going wrong, thanks
there is a mistake on the line number 29
$result1 = mysql_query("select '" . $fields . "' from `parts`") or die(mysql_error());
you are using ' instead of `. Do as follows
$result1 = mysql_query("select `" . $fields . "` from `parts`") or die(mysql_error());
Hope your problem is solved.
As it stands now, the second set of selects will be issued OUTSIDE of your </form> tag, so will never get submitted with the rest of the form. At best, you should move the form closing tag to below the POST handler.
here database details
mysql_connect('hostname', 'username', 'password');
mysql_select_db('database-name');
$sql = "SELECT username FROM userregistraton";
$result = mysql_query($sql);
echo "<select name='username'>";
while ($row = mysql_fetch_array($result)) {
echo "<option value='" . $row['username'] ."'>" . $row['username'] ."</option>";}
echo "</select>";
here username is the column of my table(userregistration)
it works perfectly