I try to insert some information to database i need to select and option then to remember the last option so i don't select all the time same option
The problem with me code is it remembers the last loop.
<form action="" method="post">
<select name="db_tablelist" onchange="showTables(this.value)">
<option value="">Select a table:</option>';
<?php
$result = mysqli_query($con,"SHOW TABLES");
while($row = mysqli_fetch_array($result)) {
if (isset($_POST['db_tablelist']) == $row[0]){
echo '<option value="' . $row[0] . '" selected="selected" >'.$row[0].'';
}
else
{
echo '<option value="' . $row[0] . '">'.$row[0].'';
echo '';
}
}
echo '</select>';
?>
Your code has problem when you are mapping post data with db data. You are checking post with isset function but you are also mapping immediately. So you have add one more condition with & operator.
This should be work.
if (isset($_POST['db_tablelist']) and $_POST['db_tablelist'] == $row[0]){
echo '<option value="' . $row[0] . '" selected="selected" >'.$row[0].'';
}
I think the problem is here:
if (isset($_POST['db_tablelist']) == $row[0])
You're checking if isset result equals to $row[0]
Related
I retrieved the selection of drop down from database. Now, I want to edit the entries. When I open the page for edit, I should have the selected data of drop down displayed in the page.
I used the following code to populate the drop down list.
<?php
try {
$sql1 = "select vehicleno from vehicle";
$projresult = $dbh->query($sql1);
$projresult->setFetchMode(PDO::FETCH_ASSOC);
} catch (PDOException $e) {
die("Some problem getting data from database !!!" . $e->getMessage());
}
echo '<select name="vehicleno" id="vehicleno" class="form-control" required value="<?php echo htmlentities($result->vehicleno);?>>';
$vno=$_POST['vehicleno'];
while ( $row = $projresult->fetch() ) {
echo '<option value="'.$row['vehicleno'].'">'.$row['vehicleno'].'</option>';
}
echo '</select>';
?>
You can't set the value of a select box directly in html, therefore as you loop through the options you'd have to check if that is the option selected. For example...
echo '<select name="vehicleno" id="vehicleno" class="form-control" required';
$vno = $_POST['vehicleno'];
while ( $row = $projresult->fetch() ) {
echo '<option value="' . $row['vehicleno'] . '"' . ($vno == $row['vehicleno'] ? 'selected' : '') . '>'.$row['vehicleno'].'</option>';
}
echo '</select>';
If this was your parent page and you wanted to send the value to a subpage such as edit-vehicle.php then you'd have to use a form element...
echo '<form action="edit-vehicle.php" method="post"><select name="vehicleno" id="vehicleno" class="form-control" required';
$vno = $_POST['vehicleno'];
while ( $row = $projresult->fetch() ) {
echo '<option value="' . $row['vehicleno'] . '"' . ($vno == $row['vehicleno'] ? 'selected' : '') . '>'.$row['vehicleno'].'</option>';
}
echo '</select></form>';
See https://www.w3schools.com/php/php_forms.asp for more about php form handling.
If I given isset() undefined error problem solved. but after submit the selected values not displayed. without isset() of how to use $_GET['ddlpub'] in php 7.2.18. it is working in php 5.2.3
<select name="ddlpublisher" id="ddlpub" class="ddltxt" >
<option value="%">--ALL--</option>
<?php
include('dbconfig.php');
$query = "Select distinct pub from jd order by pub";
$result = mysqli_query($conn, $query);
while ($row = mysqli_fetch_array($result))
{
if($row['pub']== isset($_GET['ddlpub']))
{
echo '<option value="'.$row['pub']. '" selected="selected">'.$row['pub']. '</option>';
}
else
{
echo '<option value="'.$row['pub']. '">'.$row['pub']. '</option>';
}
}
?>
</select>
You have used $_GET not post.
Also you have to check that $_GET['ddlpub'] is set and then you have to compare it as below
while ($row = mysqli_fetch_array($result))
{
if( isset($_GET['ddlpub']) && $row['pub'] == $_GET['ddlpub'])
{
echo '<option value="'.$row['pub']. '" selected="selected">'.$row['pub']. '</option>';
}
else
{
echo '<option value="'.$row['pub']. '">'.$row['pub']. '</option>';
}
}
I'm printing a select list from my database.
Here's a part of my code:
<form action="" method="post">
<div class="form-item">
<select name="specialties">
<?php
$query = "SELECT * FROM specialties";
$result = mysqli_query($connection, $query);
echo '<option value="All">' . $filterTitle . '</option>';
while($row = mysqli_fetch_assoc($result)) {
$id = $row['id'];
$title = $row['title_'. $lang];
if($_POST['specialties'] == $id) {
$selected = 'selected';
}
echo '<option value="' . $id . ' "' . $selected . '>' . $title . '</option>';
}
?>
</select>
</div>
<div class="form-item">
<input class="form-submit" type="submit" name="submit" value="<?php echo $filterSubmit; ?>">
</div>
</form>
In this form when I choose some of the options, it filters the content. That's working properly and I don't need to post that part of the code.
First when I choose some option from the select list and after submit, it filter the content, but the select option shows the All value, instead of the selected option. After that I've tried with the code above and now when I submit the form, the select value shows the last option and all the options has selected attribute.
How can I change that logic, so when I select let's say the second option, after submitting the form, the selected attribute will appear only on the second option, so the text will be the second option?
CONCLUSION:
Thank you guys for your answers. You're all correct. I gave you all +1 and I choose the Hassaan's answer, because it's very well explained.
Currently you are checking/comparing values and generating variable. You need to use else condition because you must change/reset the value of generated variable. See the example below.
if($_POST['specialties'] == $id) {
$selected = 'selected';
}
else
$selected = '';
Add an else clause to reset the selected attribute to an empty string.
if($_POST['specialties'] == $id) {
$selected = 'selected';
}
else{
$selected = '';
}
You need to make empty selected variable every time like below
while($row = mysqli_fetch_assoc($result)) {
$selected = '';
$id = $row['id'];
$title = $row['title_'. $lang];
if($_POST['specialties'] == $id) {
$selected = 'selected';
}
echo '<option value="' . $id . ' "' . $selected . '>' . $title . '</option>';
}
Hiho,
I have a little problem with showing all table names on page in form .
Code bellow:
<select name="users" onchange="showTables(this.value)">
<option value="">Select a table:</option>';
$result = mysql_list_tables($db_name);
for ($i = 0; $i < count(mysql_num_rows($result)); $i++){
echo '<option value="' . mysql_tablename($result, $i) . '">' . mysql_tablename($result, $i) . '</option>';
}
echo '</select>
</form>
<br>
<div id="tablesDb"><b>All content from selected table will be listed there.</b></div></div>';
I try with mysqli too ,and i get results but still without names of the tables and nothing can be select.
Maybe someone know how to get this work.
First, you definitely want to be using MySQLi, as the MySQL extension is deprecated. Second, you can probably get all of the data you need in a simple SELECT query like this:
select `TABLE_NAME` from information_schema.tables
Try this
$mydbname = 'database_name';
$con=mysqli_connect("localhost","my_user","my_password",$mydbname);
// Check connection
if (mysqli_connect_errno()){
echo "Failed to connect to MySQL: " . mysqli_connect_error();
exit();
}
$options = '';
// for the query you can use two of the following lines (line1+line2 OR line3+line4)
//Use line (line1+line2)
$result = mysqli_query($con,"SHOW TABLES");
$column_name ='Tables_in_'.$mydbname;
// OR (line3+line4
$result = mysqli_query($con,"SELECT TABLE_NAME AS tbl FROM information_schema.tables" WHERE TABLE_SCHEMA = \"'.$mydbname.'\"; ");
$column_name ='tbl';
while($row = mysql_fetch_array($result))
$options .= '<option value="' . $row[$column_name] . '">' . $row[$column_name] . '</option>';
echo '<select name="users" onchange="showTables(this.value)">';
echo '<option value="0">Select a table:</option>';
echo $options;
echo '</select>';
I modify code to this:
<select name="db_tablelist" onchange="showTables(this.value)">
<option value="">Select a table:</option>';
$result = mysqli_query($con,"SHOW TABLES");
while($row = mysqli_fetch_array($result)) { echo '<option value="' . $row[0] . '">'.$row[0].''; } echo '';
echo '</select>
</form>
<br>
<div id="tablesDb"><b>All content from selected table will be listed there.</b></div></div>';
And now everything works .
I have a DB table with all categories (id, category) and table with all of events (id, event, categoryID).
And on the event editing form I have the select field with all of the categories (getting from the DB). But sinse I'm developing an editing form, I need to select current category by default.
This is my select field (PHP method that gets all the categories from DB and puts them in the following order):
<option value="1">Cat1</option>
<option value="2">Cat2</option>
<option value="3">Cat3</option>
<option value="4">Cat4</option>
<option value="5">Cat5</option>
Let's say, current event is under category 3, so I need the following HTML to be generated:
<option value="1">Cat1</option>
<option value="2">Cat2</option>
<option value="3" selected>Cat3</option>
<option value="4">Cat4</option>
<option value="5">Cat5</option>
How do I achieve it with PHP, if I have the catID?
Hopefully, this question is clear enough. Sorry for my bad explanation
UPD: This is my PHP code that generates category list:
public function getCatList($conf) {
$mysqli = $this->dbConnect($conf);
// Quering...
$query = "SELECT * FROM categories";
$result = $mysqli->query($query);
while($row = mysqli_fetch_array($result)) {
echo '<option value="' . $row['id'] . '">' . $row['category'] . '</option>';
}
}
Appending to your while iteration a condition will solve this for you:
while($row = mysqli_fetch_array($result)) {
$isSelected = $row['id'] == $catID;
echo '<option '.($isSelected ? 'selected="selected"' : '').' value="' . $row['id'] . '">' . $row['category'] . '</option>';
}
You're making a comparison if the current value is the same as that stored in $catID - and store the boolean result in a variable. In the echo you're just doing a conditional forking and appending the selected attribute if the value was true, otherwise not appending any empty string.
You can do it like
while($row = mysqli_fetch_array($result)) {3
echo '<option value="' . $row['id'] . '"';
//if condition is met then make the option selected
if($row['categoryID'] == 3) {
echo " selected='selected' ";
}
echo '>' . $row['category'] . '</option>';
}
you should add the selected catID as a parameter to your getCatList-function. Then just change the creation of your HTML to include the selected-attribute:
public function getCatList($conf, $catID = 0) {
$mysqli = $this->dbConnect($conf);
// Quering...
$query = "SELECT * FROM categories";
$result = $mysqli->query($query);
while($row = mysqli_fetch_array($result)) {
echo '<option value="' . $row['id'] . '"' . ($row['id']==$catID?' selected':'') . '>' . $row['category'] . '</option>';
}
}
now you can just pass the catID to your function:
$myConf = "something";
$myCatID = 3;
getCatList($myConf, $myCatID);
//query for the categories and options
$current_value = 3;
$total_num_of_options = mysql_num_rows($your_query_result);
for($count = 1; $count <= $total_num_of_options){
echo "<option value='".$count."' ";
if($count == $current_value)
echo "selected";
echo ">cat".$count."</option>";
}