I run a script to create an order form, this is just a really small sample. I'm not so good with PHP and dynamic forms. It pulls data from mysql database.
<td>
<h3>Round Cuts</h3>
<?php while($row = mysql_fetch_array($round_cuts)){
$round_box_value = #$row["meat_names"];
$round_box_value_name = #$row["meat_names"];
echo " <input type=\"checkbox\" name=\"round_box_value\" value=\"$round_box_value_name\"> $round_box_value_name";
echo "<br>";
}?>
</td>
I've ever really only built basic contact forms, how could I process a dynamic form like this. If all else fails I would just take all the possible elements and program this like it was a not dynamic. There must be a better way though.
Even a link to a website would be helpful. I've been searching for a solution. Thanks.
I'll assume your table as a primary key of id
Start off with a minor change to your checkboxes:
<?php
while($row = mysql_fetch_array($round_cuts)){
echo sprintf('<label><input type="checkbox" name="round_box_value[]" value="%s"> %s</label><br>', $row['id'], $row['meat_names']);
}
?>
The name now has an [] at the end to tell php it's an array of values + I've wrapped the checkbox in a label for convenience.
Then when you process the form, you can simply iterate through round_box_value like so
<?php
foreach ($_POST['round_box_value'] as $id) {
// $id is the table reference from your previous table
}
// or query all the rows selected
$ids = array_map('intval', $_POST['round_box_value']); // "basic" sql injection handler
$sql = "SELECT * FROM table WHERE id IN (".implode(",", $ids).")";
should produce something like:
SELECT * FROM table WHERE id IN (2,5,7)
Related
I have project with HTML PHP and SQL database. I designed HTML input form to collect data from user and put drop down menu using PHP. It collected data from SQL database using query and those data used to drop down list and user can select those data. I have using following codes in PHP section.
$query2 = "SELECT Name FROM class ORDER BY Name";
$results2 = mysqli_query($connection, $query2);
while($result = mysqli_fetch_array($results2))
{
$Classes .= "<option value=\"{$result['Name']}\">{$result['Name']}</option>";
}
In my database there are several classes as Class_A, Class_B, Class_C and etc. (Admin can create new class or delete available class. Then when user select classes, showing classes can be different according to Admin's settings.)
In HTML form section I used following code to display Class names.
<p>
<label for=" ">Classes Are</label><br>
<select name="Classes[]" size="3" multiple >
<?php echo $Classes ?>
</select>
</p>
I tried to use this process through Check boxes (replace Drop down list from Check boxes) But I cant fix that. So, If someone can give proper way to do this it is highly appreciate.
Perhaps this is what you need to use checkboxes? The name assigned to the HTML input element uses an array type syntax with the name of from the db supplied
$sql = "select `name` from `class` order by `name`";
$res = $connection->query( $sql );
while( $rs = res->fetch_object() ) {
$classes=sprintf(
'<label>%1$s:
<input type="checkbox" value="%1$s" name="Classes[%1$s]" />
</label>',
$rs->name
);
}
Then, to process the form submission once the user has selected one or more checkboxes you could do:
foreach( $_POST['Classes'] as $index => $value ){
// do things...
}
You likely do not even need to add the name so perhaps simply Classes[] as the input name would suffice. Please clarify the issue you had when trying to use checkboxes rather than select menus.
I am busy working on a university assignment which must do the following:
Show list of sporting events for week
User Submits the teams they think will win
At end of week, calculate which users had the most correct picks
In working on above I have created a form like this, which lets the user select who they think will win.
The form data is collected from a database called events which looks like this:
My form is generate with the following code
//create form
echo'<form name="" action="" method="post">';
echo'<fieldset>';
echo'<legend>Make Your Picks</legend>';
//create query
$sql = "Select * FROM events";
$result = mysql_query($sql);
while($row=mysql_fetch_array($result)){
$team1 = $row['team1'];
$team2 = $row['team2'];
//dislay teams
echo'<h3>'.$team1.' VS '.$team2.'</h3>';
echo'<select name="'.$row['event_id'].'">';
echo'<option value="'.$row['team1'].'">'.$team1.'</option>';
echo'<option value="'.$row['team2'].'">'.$team2.'</option>';
echo'</select>';
}//while
echo'</legend></fieldset>';
echo'<input type="submit" class="buttono" value="Submit" name="submit" />';
echo'</form>'
I would like to get a more experienced users opinion on my logic and perhaps a tip if I am on the correct path and what I could be doing better, in terms of implementation,since I have ZERO confidence in my current approach.
I am creating multiple selectboxes inside the while loop, each select box has a unique name of the event_id value from db which will get inserted into the picks database (see below), this doesn't seem very efficient is there an alternative way I can approach this?
Am I correct in assigning the name attribute of the selctbox the value of the fixture_id?
Is there perhaps a more efficient way I can approach this problem that you can suggest?
Just thought it would be interesting getting an experienced users view on this
use an array for your name variable like name="event[7]" which would look like
echo'<select name="event['.$row['event_id'].']">';
then when you are gathering data you run a loop like
foreach($_POST['event'] as $event_id => $winner){
// do something here
}
I'm trying to create a form that retrieves data from a database and then allows me to add data to one column for multiple entries.
Every entry has an ID, a lot of other fields, and a category. I am trying to add these categories for every ID in the database using one form.
I came up with the solution below, but (of course)this only inserts the LAST entry in the form, because the variable ID is changed with every new row.
The form I have now shows me what I want to see, but it does not save it the way I need it to.
The question is, (how) can I make a form that has all entries in the database with a dropdown menu next to it,
lets me select the right category from the dropdown, and save it to the database?
The form:
$result = mysqli_query($con,"SELECT * FROM aw");
while($row = mysqli_fetch_array($result))
{
echo '<tr><td><input type="hidden" name="ID" value="'.$row[ID].'."> '.$row[ID].'</td><td>';
echo '
<select name="cat" onchange="this.form.submit()">
<option value="C1">category1</option>
<option value="C2"">category2</option>
</select></td></tr>
';
}
?>
<tr><td><input type="submit" title="SAVE" ></td></tr>
</form>
The insert.php
$sql="REPLACE INTO aw (ID,cat)
VALUES
('$_POST[ID]','$_POST[cat]')";
if (!mysqli_query($con,$sql))
{
die('Error: ' . mysqli_error($con));
}
mysqli_close($con);
?>
I changed my code according to Tom's answer and I now have the following:
This does print the values like they should be, but it still saves only the last entry into the database. I'm sure I must be missing something here..
$name = $_POST['ID'];
$category = $_POST['cat'];
foreach( $name as $key => $n ) {
$sql="REPLACE INTO aw (ID,cat)
VALUES
('$n','$category[$key]')";
print "The id is ".$n.", category is ".$category[$key]."<br>";
}
First of all, use PDO::Mysql, the SQL functions you are using are a bit deprecated and do not focus much on security. At the moment your code is vulnerable to SQL injections and your output is sensitive to XSS attacks (always sanitize output).
I was wrong, MySQL is deprecated but MySQLi is not! I do prefer using PDO::Mysql because of the range of databases it supports (MySQLi only supports a MySQL database, PDO::Mysql supports many more)
Now to your original question, you can create a sort of array. By making name="ID" to name="ID[]" and name="cat" to name="cat[]".
Now you can do
$name = $_POST['ID'];
$category = $_POST['cat'];
foreach( $name as $key => $n ) {
print "The id is ".$n.", category is ".$category[$key];
}
The problem is your using the name elements regardless of how many rows..
So name="ID" & name="cat" needs to change on each row or have an array type
you could use something like name="ID[]" as this would append/ create an array to $_POST['ID']... but you still would want to change your SQL query to handle each of these.
EDIT
If i understand, you want to be able to identify a row from the table so you can use that in the database?? One way todo this is when creating the table.. Give the TR a id/name attribute that is the row id from the database.
Then can simply know by checking that if your using the select menu from row #4, you check the id/name attribute of the current row and you have your database id.
<tr id='my_row_1'>
<td class='colName'>John</td>
<td class='colPhone'>1111</td>
<td class='colOther'>....</td>
</tr>
<tr id='my_row_2'>
<td class='colName'>Bill</td>
<td class='colPhone'>2222</td>
<td class='colOther'>....</td>
</tr>
<tr id='my_row_3'>
<td class='colName'>Roger</td>
<td class='colPhone'>3333</td>
<td class='colOther'>....</td>
</tr>
With something like the above, Say i had a button in on of the columns... When i click on that button.. all i have todo is find the parent TR and get its id value.... Then explode it by "_" and get the last piece to have the id..
So your PHP would generate the id easily... Also, using a form would not be the best case here.. Using multiple forms within a table is... wasteful... sort of ..
I would suggest more so, having a button that simple calls a js function which will then post/ajax/jquery what you need from that row.
--- Trying to understand exactly what you need??
Need a little help...
I have a basic html table with text field form in last column and a hidden field on each row of the web table. The data in this table is extracted out of a table in the database.
I want my users to be able to update one field in the database (a score) using this web form (page).
I have a hidden web form component on each row that contains the unique id of the record in the database for each row in the web page table.
I was attempting to create code that would update the entire list of entries on the web form, even if the user is not updating that particular field. (The values of the scores field are populated into the web form at the creation of the table. So if you did not update any scores, but hit the submit button, it would update the database table with the same values.)
Here’s my code: (abbreviated to save bytes…)
<?php
//Do all the database connection stuff up front
if (isset($_POST[‘score’]))
{
$student_id = $_POST[‘student_id’];
$score = $_POST['score'];
$n = count($score);
$i = 0;
echo "You have updated these student scores on this assignment. \r\n" .
"<ol>";
while ($i < $n)
{
echo "<hr><P>{$score[$i]} \r\n";
echo "<hr><P>{$student_id[$i]} \r\n";
$qry = "UPDATE assignments SET score = ".$score[$i]." WHERE student_id = " .$student_id[$i]. '"';
$result=#mysql_query($qry);
$i++;
}
}
if($result) {
header("location: member-index.php");
exit();
}else {
die("Query failed");
}
?>
Am I on the right track? Is there a better way to do what I’m attempting? All suggestions and ideas welcome!
Thank you in advance!
i'm guessing you are using
<input name="scores[]"/>
why not echo the unique id into the input name?
<input name="score[<?php echo $unique_id ?>]" />
this means in the loop, the $i would be the unique id (one less variable and less HTML).
and please use mysql_real_escape_string() when working with DB transactions. I know it's an example code but please don't forget that
Besides that, yes, you are on the right track
I have inserted some check box values in mysql database using PHP
And in the below image i have fetch the values:
Now i need the o/p like the below image: The values which i inserted in the database should be checked
Hope now its clear.
Thanks in advance..
You should have a table of available options (in this case, something like a cities table), and then a user-to-cities look-up table. Then you can loop over the cities, but also fetch which cities the user has checked.
A sample, without knowing your database structure, would be as follows:
$uid = $_SESSION['user']['id']; // your logged in user's ID
$cities = array();
// get an array of cities
$sql = "SELECT id, name FROM cities";
$res = mysql_query($sql);
while ($row = mysql_fetch_object($res)) {
$cities[$row->id] = $row->name;
}
// get an array of cities user has checked
$sql = "SELECT DISTINCT city_id FROM users_cities WHERE user_id = '$uid'";
$res = mysql_query($sql);
while ($row = mysql_fetch_object($res)) {
$checked[] = $row->city_id;
}
// this would be templated in a real world situation
foreach ($cities as $id => $name) {
$checked = "";
// check box if user has selected this city
if (in_array($checked, $id)) {
$checked = 'checked="checked" ';
}
echo '<input type="checkbox" name="city[]" value="'.$id.'" '.$checked.'/>';
}
If I understand you question properly, the obvious and simplest approach is that you need to fetch records from database and when producing HTML [in a loop ot something similar] check if that value exists in array to results. You haven't given us any examples of your DB structure or code, so you must figure it our yourself how to do it.
Usually, you insert the values into the database. After inserting, you should have access to the same values again. It's not clear how you set up your script, so let's assume you redirect to another script.
What you need to do is retrieve the values for the checkboxes from your database again. Then you know which are selected. This can be used to determine if your checkbox need to be checked or not.
Note:
I assume here that the result of your query is an array with
the selected Ids as a value.
I assume here that your fields are stored in the result of
some query and is basically an array
with Field Id as key and Field Name
as Value.
E.g., something like this:
<?php
// Retrieve values from database.
$resultArray = ... some code ... ;
?>
<?php foreach ($field_types as $field_name => $field_value): ?>
<input type="checkbox" name="<?php echo $field_name; ?>" value="<?php echo $field_value ?>" <?php if (in_array($field_name, $resultArray)) { echo 'checked'; }/>
<?php endforeach; ?>
This results in a checkbox which is checked, if the field_name is inside the result array (with your already checked results). Otherwise they're just rendered as unchecked checkboxes. Hope this is clear enough.