I have a form with an option menu with Staff_Names that gets its data from "Staff" table:
<select class="form-control" name="Name" id="Name">
<option value="">select staff</option>
<?php do { ?>
<option value="<?php echo $row_Staff['Staff_Name'] ?>">
<?php echo $row_Staff['Staff_Name']?></option>
<?php } while ($row_Staff = mysql_fetch_assoc($Staff));
$rows = mysql_num_rows($Staff);
if($rows > 0) {
mysql_data_seek($Categories, 0);
$row_Staff = mysql_fetch_assoc($Staff);
}
?>
</select>
The "Staff" table also has the staff member's Email. The form elements are inserted into a new table called "Training". I would like to select the Staff_Name from the option menu and insert the Staff_Name and Email in separate fields in the Training table. I have tried explode("-", $_POST['Name'] method from How to post two values in an option field? but feel there has to be a more efficient way. I don't want to continue to explode every time I need to echo in subsequent pages. Also, explode only works in this method by exploding on another page. Would like to store both fields from form when inserting into Training table. Does anyone have an efficient method?
Related
I used this code to select classes for students from the database in a dropdown. It appeared correctly. Then I when I realized that I actually want to select more than one class for a student, I added multiple. It showed as a multiple but still inserts only one value in database table.
<?php
require_once('../config.php');
$class_result = $conn->query('select * from class');
?>
<label for="exampleFormControlTextarea1">Class</label>
<select class="form-control" name="class[]" id="class" multiple>
<option value=""></option>
<?php
if ($plot_type_result->num_rows > 0) {
// output data of each row
while ($row = $plot_type_result->fetch_assoc()) {
?>
<option value="<?php echo $row["class"]; ?>">
<?php echo $row["class"]; ?>
</option>
<?php
}
}
?>
</select>
Now I want this select class to be a multi select. If I add multiple it becomes a multiple but doesn't really inserts the multiple values I select. Only selects the last selected from the database and inserts it as a single one. How can I achieve multi select and actually pass it as an array to the database with more than one value?
$classDatas= implode(", ", $datas);
You can try to save the array in a single column by converting it to text with implote.
I have two "forms" that consist of a single dropdown list each. I'm submitting both forms with javascript onchange="this.form.submit()" I need to reserve a variable, if it is set, on the second form submit. I have the form submitting to the same page and currently, it isn't passing the first variable.
<form action="<?php echo $_SERVER["PHP_SELF"]; ?>" method="GET">
<select name="show" style="width:100%;" onchange="this.form.submit()">
<option value="cond" <?php if(isset($_GET['show'])&&$_GET['show']=='cond'||!isset($_GET['show'])) { echo 'selected'; } ?>>Condensed</option>
<option value="full" <?php if(isset($_GET['show'])&&$_GET['show']=='full') { echo 'selected'; } ?>>Show All</option>
</select>
</form>
<form action="<?php echo $_SERVER["PHP_SELF"]; ?>" method="GET">
<select name="sort_slsp" style="width:100%;" onchange="this.form.submit()">
<option value="">--Choose Salesperson--</option>
<?php
$get_all_slsp = mysqli_query($lmcon, "SELECT * FROM slsps ORDER BY slsp_name");
while($row = mysqli_fetch_array($get_all_slsp)) {
echo '<option value="' . $row['slsp_id'] . '">' . $row['slsp_name'] . '</option>';
}
?>
</select>
</form>
When the form method is GET, browsers overwrite any possibly existing query string part of the URL specified via the action attribute with the new query string they construct from the form fields.
The easiest solution here is to add that additional value you want to submit as a hidden input field into the form:
<input type="hidden" name="foo" value="bar">
Seing as your first select field has the name show, and that is likely the value you want to pass on here(?), you’d fill that value attribute as such,
value="<?php echo htmlspecialchars($_GET['show']); ?>"
(Adding a check for whether that parameter exists in the first place, and maybe output a default value if not, I’ll leave up to you.)
Building a football prediction website. I am getting thee Home_team names and away team a names from fixtures table in DBMS, with corresponding drop down boxes for each fixture so that the user can predict the score. I cant get it to work. Grateful for any help!
//establish connection
<?php
$connection = mysql_connect('localhost', 'root', 'password');
mysql_select_db('mls');
$query = "SELECT * FROM fixtures WHERE Fixture_ID BETWEEN '1' and '10' ";
$result = mysql_query($query);
$num = mysql_num_rows($result);
if($num>0){
echo"<table>";
echo "<th>Home Team</th>";
echo "<th>Home Score</th>";
echo "<th>Away Score</th>";
echo "<th>Away Team</th>";
for($count=0;$count<$num; $count++){
$row = mysql_fetch_array($result);
echo"<tr>
<td>".$row['Home_Team']."</td>
<td>
<form id="myForm" method="post" action="process3.php">
<select name="Home_Score">
<select id='H".$count."'>
<option value="0">0</option>
<option value="1">1</option>
<option value="2">2</option>
<option value="3">3</option>
</select >
</td>
<td>
<form id="myForm" method="post" action="process3.php">
<select name="Home_Score">
<select id='A".$count."'>
<option value="0">0</option>
<option value="1">1</option>
<option value="2">2</option>
<option value="3">3</option>
</select>
</td>
<td>".$row['Away_Team']."</td>
</tr>";
}
echo"</table>";
<input type="submit" title="Submit the form">
</form>
}
?>
<html>
<?php
//process3.php file
<?php
include_once('db.php');
$Home_Score = $_POST['Home_Score'];
$Away_Score = $_POST['Away_Score'];
if(mysql_query("INSERT INTO user_prediction VALUES('$Home_Score', '$Away_Score')")){
$result = "Successfully Inserted";
else
$result = "Insert failed";
?>
//myscript.js file
?>
$("#sub").click(function(){
$.post( $("#myForm).attr("action"), $("#myForm:input").serializeArray(),
function (info){$("#result").html(info);});
});
$("#myForm").submit(function(){
return false;
});
While there is plenty of things that could be causing problems, notice here:
<select name="Away_Score">
//count id for unique values in dropdown
<select id='A".$count."'>
$count will literally be represented here as the string "$count", due to this code not being contained in PHP tags. Try correcting instances of code similar to this to something like below:
<?php
echo '<select name="Away_Score">';
//count id for unique values in dropdown
echo '<select id="A' . $count . '">';
?>
I see some problems there..
you switch between PHP and HTML appearently without knowing what
is what or how you start PHP and end it.
Your table is messed
up. You open a table, but in mid you just forget about it. you dont
close your td-tags or your table.
your select is messed up. You
have 2 select tags, one of which has even unescaped php-code written
in it.
your JS click-event does not trigger, because you named
it wrong.
Apart from you using mysql_-functions, which are deprecated and will stop working in newer PHP vewrsions, the list goes on...
In short: This code is a complete mess. Erasing this and starting all over with a clear head would be the best option.
To digress a bit, I honestly think that for an app of that size you should be employing some open source PHP packages. Sorry I can't comment yet so had to leave as an answer
I am populating a combo box from database, the problem is when I select an item from the combo box and try to save to another table in the database it picks the record ID column instead of the item itself.
this is the how am populating the combo box.
<label>State</label>
<select name="state" class="state" onChange="display(this.value)" width="142" style="width: 142px">
<option value="" selected="selected">-- Select State --</option>
<?php
$query="select * from tbl_state";
$query_result=mysql_query($query)or mysql_error();
while($row=mysql_fetch_array($query_result))
{
?>
<option value="<?php echo $row['id']; ?>"><?php echo $row['state_name']; ?></option>
<?php
}
?>
</select>
<div id="show_city" style="position: relative" height:5px;>
<label>LGA</label>
<select name="city" class="lga" width="142" style="width: 142px">
<option value="" selected="selected">-- Select LGA --</option>
</select>
</div>
</p>
Database connection
$state = $_POST['state'];
$city = $_POST['city'];
$sql="INSERT INTO members (state, city)
VALUES ('$state', '$city')";
NB. am using javascript to populate the combo box.
When you submit a form, only the field name along with its value will be sent. In your case, when the user select the state and press submit, the row id will be sent as you specify the row id at the option value tag.
Either you can use <option value="<?= $row['state_name'] ?>"> ... </option>" and then get the state name ini php directly:
$state = $_POST['state'];
or leave the populating code as is now and get the state name by querying database using the record id.
$state_id = intval($_POST['state']);
$city_id = intval($_POST['city']);
$sql = "SELECT `state_name` FROM tbl_state WHERE id=$state_id";
$query_result = mysql_query($sql) or mysql_error();
$state = mysql_result($query_result, 0);
echo $state;
getcity.php
<?php
$con=mysql_connect('localhost','root','') or die('Mysql not connected');
mysql_select_db('thriftdb',$con) or die('DataBase not connected');
$state_id=$_REQUEST['state_id'];
$query="select * from lga where state_id='$state_id'";
?>
<label>LGA</label>
<select name="city" width="142" style="width: 142px">
<option value="" selected="selected">-- Select LGA --</option>
<?php
$query_result=mysql_query($query)or mysql_error();
while($row=mysql_fetch_array($query_result))
{
?>
<option value="<?php echo $row['id']; ?>"><?php echo $row['lga_name']; ?></option>
<?php
}
?>
</select>
database structure is as follows
'lga' table has...
lga_id --- state_id --- lga_name
'state' table has
state_id --- state_name
When a form sends the selected item of a combo, it sends the value of the option selected, not the text. If you want to send the text, add a hidden input for every combo, then in the onclick event of the combos, call a function that populates the text to the hidden input.
selectState = function() {
$('#hiddenInputField').val($("select[name='state'] option:selected").text());
}
Then in the server side you must get the value of the input hidden field instead of the select field.
I have a problem with my javascript. First of all here is my code partitions:
<select class="inputTxt" style="width: 120px;" id="yearCombo" name="yearCombo">
<option value="0">2013</option>
<option value="1">2012</option>
<option value="2">2011</option>
</select>
function searchClicked() {
var operationField = document.getElementById("<%=FeedbackReportCtrl.PARAM_OPERATION%>");
operationField.value = "<%=FeedbackReportCtrl.OPERATION_SEARCH%>";
var yearFilter = document.getElementById("<%=FeedbackReportCtrl.PARAM_YEARFILTER%>");
yearFilter.value = document.getElementById("yearCombo").options[document.getElementById("yearCombo").selectedIndex].text;
var mainForm = document.getElementById("main");
mainForm.submit();
}
Here what goes wrong is the following;
For example, when I choose the year 2011 from the combo box and then hit the search button, it brings me the desired results;however, the selected index of the box returns back to 2013. How can I sustain my selection after search function?
The issue you have isn't a javascript one. When you submit a form you refresh the whole page, removing any client-side (user or javascript) adjustments to it.
It should be set by the php/java that is generating the page you post your form to, to set a selected="selected" or relevant, based on the value you just posted.
In php this would be
if($_POST['year'] == '2013') echo ' selected="selected"';
In java or jsp there are similar ways of doing this. Javascript itself could do the same probably.
Submitting the form refreshes the page (unless done via AJAX), thus returning to the default selected value, i.e the first one.
To overcome this you need to send along with the form the chosen year - assuming that you are self-submitting - and explicitly mark this year as the selected option.
In PHP Your code would then be something like:
<?php $year = $_POST['year']; ?>
<select class="inputTxt" style="width: 120px;" id="yearCombo" name="yearCombo">
<?php for ($i=2013;$i>2010;$i--): ?>
<option value="<?php echo $i; ?>" <?php if ($year==$i) echo "selected"; ?> >
<?php echo $i; ?>
</option>
<?php endfor; ?>
</select>