So I made a query like this
global $connection;
$query = "SELECT *
FROM streams ";
$streams_set = mysql_query($query, $connection);
confirm_query($streams_set);
in my DB there are filds
ID, UID, SID, TIME (all INT type exept time)
So I am triing to print query relult into form
<form>
<select class="multiselect" multiple="multiple" name="SIDs">
<?php
global $connection;
$query = "SELECT *
FROM streams ";
$streams_set = mysql_query($query, $connection);
confirm_query($streams_set);
$streams_count = mysql_num_rows($streams_set);
for ($count=1; $count <= $streams_count; $count++) {
echo "<option value=\"{$count}\"";
echo ">{$count}</option>";
}
?>
</select>
<br/>
<input type="submit" value="Submit Form"/>
</form>
How to print out as "option" "values" SID's from my sql query?
while ($row = mysql_fetch_array($streams_set)) {
echo '<option value="'.$row['SID'].'">'.$row['SID'].'</option>';
}
<form>
<select class="multiselect" multiple="multiple" name="SIDs">
<?php
global $connection;
$query = "SELECT *
FROM streams ";
$queryResult = mysql_query($query, $connection);
while($row = mysql_fetch_assoc($queryResult)) {
echo '<option value="'. $row["id"] .'">'. $row["title"] .'</option>';
}
?>
</select>
<br/>
<input type="submit" value="Submit Form"/>
</form>
You just have to replace the id and title index with your appropriate fields.
or mysql_fetch_object()
http://php.net/manual/en/function.mysql-fetch-object.php
Ole Jak,
take a look at PHP's mysql_fetch_array http://php.net/manual/en/function.mysql-fetch-array.php, that's what you'll want to do in a while loop :-)
Related
In my database I have 2 tables:
To insert data, I have a form that populates dropdown options from the table formulation. This is what the insert form for formulation dropdown looks like:
<?php
$formulation = '';
$query = "SELECT * FROM formulation";
$result = mysqli_query($connect, $query);
while ($row = mysqli_fetch_array($result)) {
$formulation .= '<option value="' . $row["formulationID"] . '">' . $row["formulation_name"] . '</option>';
}
?>
<select>
<option value="">Select formulation</option>
<?php echo $formulation; ?>
</select>
Now I am working on the ‘Update’ form. But my question is how can I populate the ‘Formulation’ field dropdown with the data from the formulation table (like as the insert form) but pre-selected with the existing formulation value for the name from the items table? Like this image below:
I am having problem with how I should build the form. How should I proceed with this form?
<?php
$output = array('data' => array());
$sql = "SELECT * FROM items";
$query = $connect->query($sql);
while ($row = $query->fetch_assoc()) {
$output['data'][] = array(
$row['name'],
);
}
echo json_encode($output);
?>
<form action=" " method="POST">
<div>
<label>Name</label>
<input type="text"><br>
<label>Formulation</label>
<select >
<!--What should be the codes here? -->
</select>
</div>
<button type = "submit">Save changes</button>
</form>
Thanks in advance for your suggestion.
Note: I'm not a user of mysqli so maybe there will be some error, but you will get the idea. This will not tackle the update part, just the populate part
Since you are editing a certain item, I will assume that you have something to get the item's itemID.
<?php
$sql = "SELECT * FROM items WHERE itemID = ?";
$query = $connect->prepare($sql);
$query->bind_param("s", $yourItemID);
$query->execute();
$result = $query->fetch_assoc();
$itemName = $result['name'];
$itemFormulation = $result['formulation_fk'];
//now you have the name and the formulation of that certain item
?>
<form action=" " method="POST">
<div>
<label>Name</label>
<input type="text" value="<?php echo $itemName; ?>"><br>
<label>Formulation</label>
<select >
<?php
$query = "SELECT * FROM formulation";
$result = mysqli_query($connect, $query);
while ($row = mysqli_fetch_array($result)) {
?>
<option value="<?php echo $row['formulationID']; ?>" <?php echo ($row['formulationID'] == $itemFormulation) ? 'selected' : ''; ?>>
<?php echo $row['formulation_name']; ?>
</option>
<?php
}
?>
</select>
</div>
<button type = "submit">Save changes</button>
</form>
I changed the code to better suit the problem, there may be typos, just comment for clarification
If I have understand Your question... You have to put Your result into a string. For example:
<?php
$output = array('data' => array());
$sql = "SELECT * FROM items";
$query = $connect->query($sql);
$option = '';
while ($row = $query->fetch_assoc()) {
$name=$row['name'],
$option.='<option value="$name">$name</option>'
}
echo json_encode($output);
?>
<form action=" " method="POST">
<div>
<label>Name</label>
<input type="text"><br>
<label>Formulation</label>
<select >
<?=$option?>
</select>
</div>
<button type = "submit">Save changes</button>
</form>
I hope to be of help
This should do the trick:
<?php
$itemsSql = "SELECT * FROM items WHERE itemId = 5";
$itemQuery = $connect->query($sql);
$item = $itemQuery->fetch_assoc();
$formulationsSql = "SELECT * FROM formulation";
$formulationsQuery = $connect->query($sql);
$formulations = $itemQuery->fetch_assoc();
?>
<form action="updateItem" method="POST">
<div>
<label>Item Name</label>
<input type="text" value="<?= $item[0]['name']; ?>"><br>
<label>Formulation</label>
<select>
<?php foreach($formulations as $formulation){
echo '<option value="'. $formulation['formulationId'].'">' .
$formulation['formulation_name'] . '</option>';
} ?>
</select>
</div>
<button type = "submit">Save changes</button>
</form>
I'm trying to do a select from a table based on the post value of an HTML select box. I'm getting no results at all, I'm echoing out the post value no problem. The statement works on it's own but won't when I use the select form to populate it. This is just my test I will be adding other options to the dropdown box.
<?php
if(isset($_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);
?>
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>";
I need to have about 60 different fieldsets within a form that all contain the same elements. The only difference is the ID being called by my query. The ID will be populating data for me from mysql.
Is there a way to do this more efficiently other than adding 60 fieldsets?
<fieldset>
<h2>Make Your Pick</h2>
<?php
require_once ('mysql_connect.php');
$query = "SELECT id, name FROM table WHERE id = 1";
$result = #mysql_query ($query) or die ('error submitting' .mysql_error());
echo "<select name='winner' id='winner'><option>Who Will Win?</option>";
while($drop=mysql_fetch_array($result)){
//data stored in $drop
echo "<option value=$drop[id]>$drop[name]</option>";
}
echo "</select>";
?>
<input id="insert" type="submit" value="Next" />
</fieldset>
I found the answer!
<?php
for ($i = 1; $i <= 60; $i++) {
echo '<fieldset>
<h2 class="fs-title">Make Your Pick</h2>';
require_once ('mysql_connect.php');
$query = "SELECT id, name FROM table WHERE id = $i";
$result = #mysql_query ($query) or die ('error submitting' .mysql_error());
echo "<select name='winner' id='winner'><option>Who Will Win?</option>";
while($drop=mysql_fetch_array($result)){
//data stored in $drop
echo "<option value=$drop[id]>$drop[name]</option>";
}
echo "</select>";
echo '<hr>
<input id="insert" type="button" name="next" class="next action-button" value="Next" />
</fieldset>';
}
?>
I have this code and I'm trying to put the selected state in a subcat table.
So far it returns an empty value. I'm not sure if this is clear or not, but all I want is: select a state from the select option and submit it. I want to get the selected state name into my table subcat.
enter <?php
include("connect.php");
$state = $row['states']; //Select name
if (isset($_POST[submit])){
$query = "INSERT INTO subcat (sub_name) VALUES ('$state')";
mysql_query($query) or die(mysql_error());
}
?>
<form action="" method="post" name="form">
<?php
$sql = mysql_query("SELECT * FROM state");
echo "<select name='states'>
<option value=''>Select a state</option>";
while ($row = mysql_fetch_assoc($sql)) {
echo "<option value='$row[id]'>$row[name]</option>";
}
echo "</select>";
?>
<input type="submit" name="submit" value="Continue" />
</form> here
Thanks
Change $state = $row['states'] to $state = $_POST['states']
<?php
include("connect.php");
if (isset($_POST[submit]))
{
$state = $_POST['states']; //Select name
$query = "INSERT INTO subcat (sub_name) VALUES ('$state')";
mysql_query($query) or die(mysql_error());
}
?>
<form action="" method="post" name="form">
<?php
$sql = mysql_query("SELECT * FROM state");
echo "<select name='states'>
<option value=''>Select a state</option>";
while ($row = mysql_fetch_assoc($sql)) {
echo "<option value='$row[id]'>$row[name]</option>"; // if you want to
//get the name into table, then use like this
//echo "<option value='$row[name]'>$row[name]</option>"; or
//echo "<option>$row[name]</option>";
}
echo "</select>";
?>
<input type="submit" name="submit" value="Continue" />
</form>
Try this:
enter <?php
include("connect.php");
if (isset($_POST[submit])){
$state = $_POST['states'];
$query = "INSERT INTO subcat (sub_name) VALUES ('".mysql_real_escape_string($state)."')";
mysql_query($query) or die(mysql_error());
}
?>
<form action="" method="post" name="form">
<?php
$sql = mysql_query("SELECT * FROM state");
echo '<select name="states" id="states">
<option value="">Select a state</option>';
while ($row = mysql_fetch_assoc($sql)) {
echo '<option value="'.$row['name'].'">'.$row['name'].'</option>';
}
echo '</select>';
?>
<input type="submit" name="submit" value="Continue" />
</form> here
Dont forget to use mysql_real_escape_string to prevent SQL injections. I have replaced $state = $row['states']; with $state = $_POST['states'];
I dont know where u got $row from...
The above will insert the states name into the database.