I've got a drop down select box that grabs each relevant value from an SQL database in a loop.
I'm creating a form so that when the "Submit" button is pressed it redirects to a PHP file that carries out the INSERT SQL statement. However because the select options are coming from a loop I'm unsure of how to grab the right value when its selected as it just grabs the last value gained from the loop.
I'm pretty sure that the way I have done it is the wrong way to go
<?php
echo"<select name='ModuleTitle' id='ModuleTitle' style='width:100%;'>";
echo"<option>Select...</option>";
//3. Perform database query
$result = mysql_query("SELECT * FROM Module
ORDER BY `ModTitle` ASC;", $connection);
if(!$result){
die("Database query failed: " . mysql_error());
}
//4. Use Returned Data
while ($row5 = mysql_fetch_array($result)) {
$module = $row5[2];
echo "<option name='{$module}'>".$row5[2]."</option><br />";
}
echo"</select>";
echo "<a href='submitREQ.php?id={$module}'><img src='images/submit.jpg' height='27'></a>";
?>
Instead of using <a href you should use <input type="image" value="submit" src="images/submit.jpg" />
To grab the value after the form is submitted you should use: $ModuleTitle = $_POST['ModuleTitle']; or $_GET if the method is get.
Related
I am creating a simple custom search form. This form searches the table 'oc_product' where i created a new column 'gcode'. I have inserted a sample data in this column for one of the products. Is it necessary to make a new db/mysql connect in php within a new tpl file i created just like 'information/contact'. I call this 'gcode/gcode'.
I tried but unable to get result form the search. It redirected to index.php.
My code is:
<form method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>" id="searchform">
<input type="text" name="name">
<input type="submit" name="submit" value="Search">
</form>
if(isset($_POST['submit'])){
if(isset($_GET['go'])){
if(preg_match("/^[ a-zA-Z]+/", $_POST['name'])){
$name=$_POST['name'];
//connect to the database
$db=mysql_connect ("localhost", "username_demo", "pwddemo123") or die ('I cannot connect to the database because: ' . mysql_error());
//-select the database to use
$mydb=mysql_select_db("db_demo");
//-query the database table
$sql="SELECT * FROM oc_product WHERE gcode LIKE '%" . $name . "%'";
//-run the query against the mysql query function
$result=mysql_query($sql);
//-create while loop and loop through result set
while($row=mysql_fetch_array($result)){
$gcode =$row['gcode'];
//-display the result of the array
echo '<span>'."This product is genuine".$gcode.'</span>';
}
}
else{
echo "<p>Please enter a search query</p>";
}
}
}
Any example to search a column and fetch data from a column.
Don't quite understand your question. Do you want to fetch data from ONE column? Also use mysql_fetch_assoc instead mysql_fetch_array
And replace $_GET with $_POST
Also your select statement is wrong, so do this:
<?
$sql="SELECT * FROM oc_product WHERE gcode LIKE '%$name%'";
$result = mysql_query($sql);
$results = mysql_num_rows($result);
for($i=0; $i<$results; $i++)
{
$row = mysql_fetch_assoc($result);
$gcode = $row['gcode'];
// Here list it the way you want
echo $gcode.'<br>';
}
?>
you would need to create a model or simply add this field in the search model to be searched by it. i am not sure if that is what you want to accomplish, also you would want to make sure to sanitize the request and escape it before you get one of the SQL injection attacks.
I am having trouble in getting checked values checked in form. I was trying to use the same function as I have used to insert values to print all values in edit form, and which are in other table inserted to mark them checked.
Function to insert values in database table in insert form and it works.
function emarketing_usluge(){
$link = new mysqli("localhost", "xxx", "xxx", "xxx");
$link->set_charset("utf8");
$sql=mysqli_query($link, "SELECT * FROM `jos_ib_emarketing_oprema` order by OpremaId asc ");
while($record = mysqli_fetch_array($sql)) {
echo '<input type="checkbox" name="usluge[]" value="'.$record['OpremaId ']. '">' . $record['OpremaNaziv'] . ' <br/><br/> </input>';
}
}
In this function I get list of all services and place them in checkboxes.
Now I want to edit form, and display all values that are checked by using same function.
First I make query to get values, I am using here pdo but for funcion files I have used mysqli.
Form for editing!
$sql_oprema = "SELECT a.Partner, a.OpremaId, a.Oprema, b.OpremaNaziv
FROM jos_ib_emarketing_stavke_oprema a
join jos_ib_emarketing_oprema b
on OpremaId = b.Oprema
WHERE a.Partner= $id";
$oprema = $conn->query($sql_oprema);
$row = $oprema ->fetch();
<div class="col-xs-6">
<input type="checkbox" id="oprema" onclick="Exposeoprema()">Oprema<br>
<div id="Scrolloprema" style="height:150;width:200px;overflow:auto;border:1px solid blue;display:none">
<?php
while($row = $oprema ->fetch()) {
$data='<input type="checkbox" name="oprema[]" value="'.$row["Oprema"].'"';
if(isset($row['Oprema'])) {//field in the database
$data.=' checked="checked';
}
$data.='">'. $row["OpremaNaziv"] .'</br>';
}
emarketing_oprema($data);
?>
</div>
</div>
I am trying print all service values by using function, but the ones that are checked they need to have check mark. I am getting problem and could not figure it out how to solve it.
Looking back to your SQL query, I don't see an extraction of checked field, you are not selecting it. So there is never going to be a $row['checked'] element of your query.
You should add:
$sql_oprema = "SELECT a.checked, a.Partner, a.OpremaId, a.Oprema, b.OpremaNaziv
FROM jos_ib_emarketing_stavke_oprema a
join jos_ib_emarketing_oprema b
on OpremaId = b.Oprema
WHERE a.Partner= $id";
<?php
// select box open tag
$selectBoxOpen = "<select name='store_name' >";
// select box close tag
$selectBoxClose = "</select>";
// select box option tag
$selectBoxOption = '';
// connect mysql server
$con = mysql_connect("localhost","root","");
if (!$con) {
die('Could not connect: ' . mysql_error());
}
// select database
mysql_select_db("store", $con);
// fire mysql query
$result = mysql_query("SELECT store_name FROM store_input");
// play with return result array
while($row = mysql_fetch_array($result)){
$selectBoxOption .="<option value = '".$row['store_name']."'>".$row['store_name'] . "</option>";
}
// create select box tag with mysql result
$selectBox = $selectBoxOpen.$selectBoxOption.
$selectBoxClose;
echo $selectBox;
?>
this is my sample code, I have created combobox in php with option values from database values
but i'm not able to change the page contents when i select options.
any answers
Your code is lacking Javascript, which would be required for the outcome you are expecting. Seeing as you haven't really defined the complete outcome you expect, I will just show you what you need to update the url with the currently selected store:
$selectBoxOpen = "<select name='store_name' onchange=\"location.href=location.href+'?store='+this.value\" >";
I've added an onchange event (read more about DOM events here) so that when you select a different value you are changing the location.
Now when you change the value of the select, your url should change to scriptname.php?store=SomeStoreName
So, im creating a form that will submit data into an SQL database. Ive got 2 select drop downs that hold the data of "Module Code" and "Module Title". In the database a module title will only have one module code e.g Team project(module title) has module code 11COB290.
How can i get it so that when a user selects a given module name OR Module title is will automatically select the correct partner i.e select the right module code thats related to the module name the user has selected without pressing any submit buttons?
The following code is the drop down select boxes and php code i have so far:
<td align="center">
<select name='ModuleTitle' id='ModuleTitle' style='width:100%;'>
<option>Select...</option>
<?php
//3. Perform database query
$result = mysql_query("SELECT * FROM Module
ORDER BY `ModTitle` ASC;", $connection);
if(!$result){
die("Database query failed: " . mysql_error());
}
//4. Use Returned Data
while ($row = mysql_fetch_array($result)) {
$module = $row[2];
echo "<option name='{$module}'>{$module}</option><br />";
}
?>
</select>
</td>
<td align="center">
<select name='ModuleCode' id='ModuleCode' style='width:100%;'>
<option>Select...</option>
<?php
//3. Perform database query
$result = mysql_query("SELECT * FROM Module
ORDER BY `ModCode` ASC;", $connection);
if(!$result){
die("Database query failed: " . mysql_error());
}
//4. Use Returned Data
while ($row = mysql_fetch_array($result)) {
$module = $row[3];
echo "<option name='{$module}'>{$module}</option><br />";
}
?>
</select>
</td>
Unless there is a special reason to do so, why not give a single SELECT box instead of two?
<td align="center">
<select name='ModuleTitle' id='ModuleTitle' style='width:100%;'>
<option>Select...</option>
<?php
//3. Perform database query
$result = mysql_query("SELECT * FROM Module ORDER BY `ModTitle` ASC;", $connection);
if(!$result){
die("Database query failed: " . mysql_error());
}
//4. Use Returned Data
while ($row = mysql_fetch_array($result)) {
$module = $row[2] . ' (' . $row[3] . ')';
$moduleCode = $row[3];
echo "<option value='{$moduleCode}'>{$module}</option>";
}
?>
</select>
</td>
Or otherwise if you would like to keep 2 SELECTs, use AJAX calls. The idea is to define onChange event on each SELECT and in that event, send an AJAX request to a PHP script. The PHP script will request the module code or title and send it back to the AJAX handler. The AJAX handler can then automatically mark as SELECTED the corresponding option in the other SELECT. You might need sometime to research on AJAX and JavaScript if you aren't already experienced with this stuff.
Another idea might be to use the module code as the value for the title SELECT and module title as the value for the code SELECT. For example, title SELECT will be:
<SELECT name="ModuleTitle" id="ModuleTitle" style="width:100%;">
<OPTION>Select...</option>
<OPTION value="123">Title 123</OPTION>
<OPTION value="345">Title 345</OPTION>
<OPTION value="567">Title 567</OPTION>
</SELECT>
Then in the onChange event handler you might do something like this:
var selObj = document.getElementById('ModuleTitle');
var selIndex = selObj.selectedIndex;
document.getElementById('ModuleCode').value = selObj.option[selIndex].text;
By the way, there is no "name" attribute for . It should be "value" instead. Please fix in your code.
Hope it helps!
CAUTION: none of the above code is tested but I hope it works fine
I'm currently using php to populate a form with selections from a database. The user chooses options in a select style form and submits this, which updates a summary of the selections below the form before a second submit button is used to complete the interaction.
My issue is that every time a user uses the first submit, the selections that were there previously do not stick. They have to go through the whole form again.
Is there anyway to keep these selections present without resorting to php if statements? There are a ton of options so it would be a pain to use php for each one. Also, form is being submitted via POST.
Sample from form:
<?php
// GRAB DATA
$result = mysql_query("SELECT * FROM special2 WHERE cat = 'COLOR' ORDER BY cat")
or die(mysql_error());
echo "<div id='color'><select id='color' name='product_color'>";
while($row = mysql_fetch_array( $result )) {
$name= $row["name"];
$cat= $row["cat"];
$price= $row["price"];
echo "<option value='";echo $name;echo"'>";echo $name;echo" ($$price)</option>";}
echo "</select>";
echo "<input type='hidden' name='amount_color' value='";echo $price;echo"'></div>";
?>
I tried using this js snippet to repopulate the selections, but it does not seem to work properly...
<script type="text/javascript">document.getElementById('color').value = "<?php echo $_GET['proudct_cpu'];?>";</script>
This does not seem to work. Any suggestions other than php if statements?
Thanks!
edit: This is basically the form set up I'm using, though I've shortened it significantly because the actual implementation is quite long.
// Make a MySQL Connection
<?php mysql_connect("localhost", "kp_dbl", "mastermaster") or die(mysql_error());
mysql_select_db("kp_db") or die(mysql_error());
?>
<br />
<form action="build22.php" method="post">
<input type="hidden" name="data" value="1" />
<br />
<br />
<?php
// GRAB DATA
$result = mysql_query("SELECT * FROM special2 WHERE cat = 'color' ORDER BY cat")
or die(mysql_error());
echo "<div id='color'><select id='color' name='product_color'>";
while($row = mysql_fetch_array( $result )) {
$name= $row["name"];
$cat= $row["cat"];
$price= $row["price"];
echo "<option value='";echo $name;echo"'>";echo $name;echo" ($$price)</option>";}
echo "</select>";
echo "<input type='hidden' name='amount_color' value='";echo $price;echo"'></div>";
?>
<input type="submit" value="Update Configuration">
</form>
The selections from the form above get echoed after submission to provide the user with an update as such:
<div id="config" style="background-color:#FFF; font-size:12px; line-height:22px;">
<h1>Current Configuration:</h1>
<?php echo "<strong>Color:</strong>    ";echo $_POST['product_color']; ?>
</div>
I assume you're storing the user's selections in a separate table. If that's the case, you'll need to add some logic to determine if you should display the form values or what's already been stored.
<?php
// form was not submitted and a config id was passed to the page
if (true === empty($_POST) && true === isset($_GET['config_id']))
{
// make sure to properly sanitize the user-input!
$rs = mysql_query("select * from saved_configuration where config_id={$_GET['config_id']}"); // make sure to properly sanitize the user-input!
$_POST = mysql_fetch_array($rs,MYSQL_ASSOC); // assuming a single row for simplicity. Storing in _POST for easy display later
}
?>
<div id="config" style="background-color:#FFF; font-size:12px; line-height:22px;">
<h1>Current Configuration:</h1>
<?php echo "<strong>Color:</strong>    ";echo $_POST['product_color']; ?>
</div>
So after storing the user's selections in the database, you can redirect them to the page with the new config_id in the URL to load the saved values. If you're not storing the selected values in a table, you can do something similar with cookies/sessions.
echo the variables into the value tag of the form elements. If you post all your code I'm sure I can help you.
UPDATE
ah, so they are dropdown lists that you need to remember what was selected? Apologies, I read your post in a rush yesterday and thought it was a form with text inputs.
I just did a similar thing myself but without trying your code let me see if I can help.
Basically what you need to do is set one value in the dropdown to selected="selected"
When I had to do this I had my dropdown values in an array like so:
$options = array( "stack", "overflow", "some", "random", "words");
// then you will take your GET variable:
$key = array_search($_GET['variablename'], $options);
// so this is saying find the index in the array of the value I just told you
// then you can set the value of the dropdown to this index of the array:
$selectedoption = $options[$key];
This is where it might be confusing as my code is different so if you want to use it you will probably need to restructure a bit
I have a doSelect function to which I pass the following parameters:
// what we are passing is: name of select, size, the array of values to use and the
// value we want to use as the default selected value
doSelect("select_name", 1, $options, $selectedoption, "");
// these are the two functions I have:
// this one just processes each value in the array as a select option which is either
// the selected value or just a 'normal' select value
FUNCTION doOptions($options, $selected)
{
foreach ($options as $option)
{
if ($option == $selected)
echo ("<option title=\"$title\" id=\"$value\" selected>$option</option>\n");
else
echo ("<option title=\"$title\" id=\"$value\">$option</option>\n");
}
}
// this is the function that controls everything - it takes your parameters and calls
// the above function
FUNCTION doSelect($name, $size, $options, $selected, $extra)
{
echo("<select class=\"\" id=\"$name\" name=\"$name\" size=\"$size\" $extra>\n");
doOptions($options, $selected);
echo("</select>\n");
}
I know that's a lot of new code that's been threw at you but if you can get your select values from the db into the array then everything else should fall nicely into place.
The only thing I would add, is at the start where we call doSelect, I would put that in an if statement because you don't want to set something as selected which hasn't been set:
if (isset($_GET['variable']))
{
$key = array_search($_GET['variablename'], $options);
$selectedoption = $options[$key];
doSelect("select_name", 1, $options, $selectedoption, "");
}
else
{
doSelect("select_name", 1, $options, "", "");
}
I hope that helps!