I would like some help printing what is in a certain textbox that was created by an echo command.
while($row = $result->fetch_assoc()){
$stringTest = $row['Price'];
$AssetId = $row['AssetId'];
echo "<center><div> <h3>Cost: ".$stringTest."";
echo '<form action="" method="get"><input type="text" name="uid">';
echo "</br><input class='myButton' type='submit' Name='Submit1' VALUE='I have bought'></a></form>";
/** ^ Input value I would like to get *//
echo "<a href='https://www.roblox.com/item-item?id=".$AssetId."' class='myButton'>Buy</a></h3></div></center>";
}
Use the code below to get the value from submit:
if(isset($_GET['Submit1'])) {
echo $_GET['Submit1'];
}
When the user clicks submit, it will echo the value of it.
If you want to print PHP element in a textbox you should put it in the value tag of the input
<?php
echo "<input type='text' value='" . $val . "'>";
?>
Related
I am trying to update a php form that holds a few rows of mysql data. I have a button next to each row and when i click on that I want to update the row. The issue im having below is the ID is only set as the last row. How do i get this to push the ID to the button? So basically no matter what button i press i always get the same ID which is the last one to load.
if($result){
while($row = mysqli_fetch_array($result)){
$id = $row["ID"];
$beername = $row["BeerName"];
$beertype = $row["BeerType"];
$beerpercent = $row["BeerPercent"];
$beerdescription = $row["BeerDescription"];
$nowpouring = $row["NowPouring"] =='0' ? '' : 'checked=\"checked\"';
$glutenreduced = $row["GlutenReduced"] =='0' ? '' : 'checked=\"checked\"';
$beertogo = $row["BeerToGo"] =='0' ? '' : 'checked=\"checked\"';
echo "<form action='' method='POST'>";
echo "<tr><td><h6><input type=\"text\" size=\"5\" name=\"id\" value=\"$id\"></h6></td>";
echo "<td><h6><input type=\"text\" size=\"30\" name=\"BeerName\" value=\"$beername\"></h6></td>";
echo "<td><h6><input type=\"text\" size=\"30\" name=\"BeerType\" value=\"$beertype\"></h6></td>";
echo "<td><h6><textarea size=\"90\" style=\"width:250px;height:150px;\" name=\"BeerDescription\" value=\"\">$beerdescription</textarea></h6></td>";
echo "<td><h6><input type=\"text\" size=\"5\" name=\"Percent\" value=\"$beerpercent\"></h6></td>";
echo "<td><h6><input type=\"checkbox\" name=\"NowPouring\" value=\"true\" $nowpouring></h6></td>";
echo "<td><h6><input type=\"checkbox\" name=\"GlutenReduced\" value=\"true\" $glutenreduced></h6></td>";
echo "<td><h6><input type=\"checkbox\" name=\"BeerToGo\" value=\"true\" $beertogo></h6></td>";
#echo "<td><h6> <a href=\". $_SERVER["PHP_SELF"] .?id=".mysql_result($result,$j,'id')."\" onclick=\"\"></h6></td>";
echo "<td><h6> <button name=\"submit\" type=\"submit\" value=\"$id\">Save</button></h6></td>";
echo "</tr>";
echo "</form>";
}
}
if (isset($_POST['submit'])) {
$user = $_POST['submit'];
echo "<p style=\"color:#ffffff\">$id</p>";
#$delet_query = mysqli_query($mysqli, "UPDATE NowPouring SET NowPouring = '1' WHERE ID = '4'") or die(mysql_error());
if ($delet_query) {
echo '<p style="color:#ffffff">Beer with id '.$id.' is updated. To refresh your page, click ' . ' <a href=' . $_SERVER["PHP_SELF"] . ' > here </a></p>';
}
}
?>
The main problem I see here is that the while loop your code has is generating the same name for the inputs...
All of your "<button name=\"submit\" type=\"submit\" value=\"$id\">Save</button>" will have the same name, that's why it always has the last id as value.
Maybe you should try something such as..
<button name=\"$id_submit\" type=\"submit\" value=\"$id\">Save</button>
or if you want you can store it in an array..
<button name=\"submit[]\" type=\"submit\" value=\"$id\">Save</button>
You are seeing this result because the 'name' of each of your inputs is the same, so essentially you have a form with a bunch of elements that have the same names. You need to add a dynamic aspect to each name.
For example, you could update your output to something like this:
echo "<tr><td><h6><input type=\"text\" size=\"5\" name=\"id_$id\" value=\"$id\"></h6></td>";
Where each line adds the current id. Then when you retrieve the form data, you can append the submitted id to the field you want to update.
Have you considered using an AJAX approach so you can submit just the line in question and not have to reload the page and return the whole data set each time?
Make <form> for each submit button. Adding <form> in the while():
if($result){
while($row = mysqli_fetch_array($result)){
echo "<form action='' method='POST'>";
//...
echo "</form>";
}
}
Your form tag is placed at the wrong place.
It should be within:
while($row = mysqli_fetch_array($result)){
$id = $row["ID"];
//....
//....
echo "<form action='' method='POST'>";
echo"<tr>";
echo "<td>" . $id . "</td>";
//....
//....
echo "<td><button type='submit' name='submit'>Save</button></td>";
echo"</tr>";
echo "</form>";
}
I got a select box where i can choose a Project-ID and pass it to the next form with a submit button. Right now it looks like this:
This is my code:
<form name="form2" action="formular3.php" method="post">
<p><strong>Choose Project:</strong></p>
<select name = "project_id">
<?php
while ($row = mysqli_fetch_array($result)) {
echo "<option value =" . $row['project_id'] . "> Project_ID: " . $row['project_id'] . " - (" . $row['name'] . ")</option>";
}
?>
</select>
<input type="submit" value="Send" />
</form>
Now I want to have a list which has a Submit-button for every row. It should look like this:
If you are only sending one option per form, then remove the select tag entirely.
Better Method (and notice mysqli_fetch_assoc() instead of mysqli_fecth_array():
echo "<form name=\"form2\" action=\"formular3.php\" method=\"post\">";
while($row=mysqli_fetch_assoc($result)){
echo "Project_ID: {$row['project_id']} - ({$row['name']}) ";
echo "<button name=\"project_id\" value=\"{$row['project_id']}\">Send</button>";
}
echo "</form>";
This will submit the value in the clicked button without having to write so many form blocks into the html. You will only need to adjust the actual displaying of the buttons with <br>, table cells, etc.
My previous method that will work, but is not DRY:
while($row=mysqli_fetch_array($result)){
echo "<form name=\"form2\" action=\"formular3.php\" method=\"post\">";
echo "<input type=\"hidden\" name=\"project_id\" value=\"{$row['project_id']}\">";
echo "Project_ID: {$row['project_id']} - ({$row['name']}) ";
echo "<input type=\"submit\" value=\"Send\" />";
echo "</form>";
}
Use submit button inside the loop,
<p><strong>Choose Project:</strong></p>
<select name = "project_id">
<?php
while ($row = mysqli_fetch_array($result)) {
echo "<option value =" . $row['project_id'] . "> <input type='submit' value='Send' />Project_ID: " . $row['project_id'] . " - (" . $row['name'] . ")</option>";
}
?>
</select>
</form>
You need do stop occuring current submit event and give submit button to a id, use jquery to click on submit you should post a spicific form.
like on of the submit will be clicked, you will run this script.
$(document).on('click', '#submit', function () {
$(this).closest("form").submit();
});
I try to figure out how to handle two nested forms. The first one sends data from select-option dropdown to another_file.php. The nested one should send data from check boxes to current file and should be handled by isset($_POST['...']). Here is my simplified code:
if (isset($_POST['a']))
{
// do something
}
if (isset($_POST['b']))
{
// do something else
}
echo "<form action='another_file.php' method='post'>"; // begin of first form
echo "<table class ='table table-hover table-condensed table-striped table-bordered'>";
echo "<thead>";
echo "<th>ID</th><th>Jídlo</th><th>Množství</th><th>por_cislo</th><th>Odebrat</th>";
echo "</thead>";
while($rows = $stmt->fetch()){
echo "<tr><td>" . $rows['id'] . "</td><td>".$rows['jidlo'];
echo "<select name =".$rows['id']."_".$rows['por_cislo']."> ";
$stmt2 = $db->query($q2);
echo "<option value ='nic'> (vyberte potravinu) </option>";
$max_por_cislo = $rows['por_cislo'];
while($rows2 = $stmt2->fetch())
{
echo '<option value="'.$rows2['id'].'">'.$rows2['jidlo'].'</option>';
};
echo "</select>";
echo "</td> <td>" . $rows['mnozstvi'] . "g <input name = '".$rows['por_cislo']."' type='text' value = '-'></td><td>".$rows['por_cislo']."</td>";
echo "<td><form action ='this_file.php' method = 'post'>"; // begin of nested form
echo "<input type='checkbox' id='atur_peg' name='idecko[]' value=".$rows['id']."*".$rows['por_cislo']." /></td></tr> ";
};
echo "<input type='hidden' name='pc' value=".$plan_cislo.">";
echo "<tr><td colspan='2'><input name = 'go' type='submit' value='OK'/></td><td colspan='2'><input type = 'submit' name ='a' value='ADD'/></td>";
echo "<td><input type = 'submit' name ='b' value='DELETE checked'/></form></td></tr>"; // end of nested form
echo "</table>";
echo "</form>"; // end of first form
Is
there any way to do this correctly?
Instead of nested forms, use jquery to detect checkbox being checked and then change the form action url. Something on this line.. You will have to modify the same to suit your needs.
$(document).ready(function(){
$("#formname").on("change", "input:checkbox", function(){
if( $(this).is(":checked") ) {
$('#formName').attr('action', 'this_file.php');
$("#formname").submit();
}
});
});
I have a form with 2 selects, when you send the first, the second select charges the values that are called on my oracle bd with a query, then when i send the second select, it generates a table with checkboxes:
if(isset($idTActi)){
$stallTableTarifas=oci_parse($conn, "SELECT TARIFAS.ID, TARIFAS.ID_TIPO_ACTIVIDAD, TARIFAS.TIPO, TIPO_ACTIVIDAD.TEMPS_KM, TARIFAS.PRECIO
FROM TARIFAS, TIPO_ACTIVIDAD
WHERE TARIFAS.ID_TIPO_ACTIVIDAD = TIPO_ACTIVIDAD.ID
AND TARIFAS.ID_TIPO_ACTIVIDAD = $idTActi");
oci_execute($stallTableTarifas);
echo "<div class='divPrecios'>";
echo "<table>";
echo "<tr class='tabPreciosTitles'>";
echo "<td>Tipus Tarifa</td>
<td>Temps/Km</td>
<td>Preu</td>
<td><input type='submit' class='carrito' value=''></td>";
echo "</tr>";
while (($row=oci_fetch_array($stallTableTarifas,OCI_BOTH))!=false){
echo "<tr>";
echo "<td>".$row['TIPO']."</td>";
echo "<td>".$row['TEMPS_KM']."</td>";
echo "<td>".$row['PRECIO']."</td>";
echo "<td><input type='checkbox' name='checkbox[]' value='".$row['ID']."'/></td>";
echo "</tr>";
}
echo "</table>";
echo "</div>";
}
echo "</form>";
The variable $idTActi it's the id that i return from the second select, so when i click on the checkboxes and i send it on the button named class='carrito', that's an sprite that i generate on css, i see on the bottom another table with the information that i selected on the previous table:
echo "<div class='divPrecios'>";
echo "<table>";
echo "<tr class='tabPreciosTitles'>";
echo "<td>Nom Activitat</td>
<td>Nom Tipus Activitat</td>
<td>Tipus Tarifa</td>
<td>Temps/km</td>
<td>Preu</td>";
echo "</tr>";
foreach($_POST['checkbox'] as $item){
$stallTableCarrito=oci_parse($conn, "SELECT ACTIVIDAD.NOM AS NOM_ACTIVIDAD, TIPO_ACTIVIDAD.NOM AS NOM_TACTIVIDAD, TARIFAS.TIPO, TIPO_ACTIVIDAD.TEMPS_KM, TARIFAS.PRECIO
FROM TARIFAS, ACTIVIDAD, TIPO_ACTIVIDAD
WHERE TARIFAS.ID = $item
AND TARIFAS.ID_TIPO_ACTIVIDAD = TIPO_ACTIVIDAD.ID
AND TIPO_ACTIVIDAD.ID_ACTIVIDAD = ACTIVIDAD.ID");
oci_execute($stallTableCarrito);
$array=array(
0=>array(),
1=>array(),
2=>array(),
3=>array(),
4=>array()
);
while (($row=oci_fetch_array($stallTableCarrito,OCI_BOTH))!=false){
array_push($array[0],$row['NOM_ACTIVIDAD']);
array_push($array[1],$row['NOM_TACTIVIDAD']);
array_push($array[2],$row['TIPO']);
array_push($array[3],$row['TEMPS_KM']);
array_push($array[4],$row['PRECIO']);
}
for ($x=0;$x<count($array[0]);$x++){
echo "<tr>";
echo " <td>".$array[0][$x]."</td>";
echo " <td>".$array[1][$x]."</td>";
echo " <td>".$array[2][$x]."</td>";
echo " <td>".$array[3][$x]."</td>";
echo " <td>".$array[4][$x]."</td>";
echo " <td><input type='submit' class='carritoElim' value=''></td>";
echo "</tr>";
}
}
echo "</table>";
echo "</div>";
Basically that's a shopping form.
And where is the problem? When i send the pushed checkboxes with the button class='carrito', the form by default refresh the page and clears my array, what can i do?
In your first part of code, is your form tag open ? (I guess it is if this one works)
In the second part, is your <input type='submit' class='carritoElim' value=''> tag in a form ?
Because if it's not, you gonna have a bad time ;-)
Maybe in the last form you should generate hidden input with same names as your first form and same values.
If you don't I guess your variable $idTActi won't be set anymore and it won't succeed the first test if(isset($idTActi)). That could be why you get a cleared page.
If you have a multi step form in the same php page, for this kind of html code :
<form method=POST url="myURL">
<select name="select1">[...]</select>
<select name="select2">[...]</select>
<!-- VARIOUS PART : may not be displayed -->
<div id="checkboxes">
<input type="hidden" name="boxStep" value="1"/>
<input type="checkbox" name="cb1" value="1"/>
[...]
</div>
<!-- END OF VARIOUS PART -->
</form>
Then you need php tests in this order :
// if post request
if (isset($_POST)) {
if (isset($_POST['boxStep'])) {
// behavior when checkboxes values are sent
} else {
if (isset($_POST['select2'])) {
// behavior when second select is filled
// display "VARIOUS PART"
} else {
// behavior when only first select is filled
// Do not display "VARIOUS PART"
}
}
} else {
// default behavior (no select filled)
// Do not display "VARIOUS PART"
}
Apolo
I have a php page with this code that passes a variable of a button to next page:
<div><center><table>
while($row = mysqli_fetch_array($result))
{
echo "<td><form action= display.php method= 'post'><input type='hidden'
name='projectid' value=".$row['projectid'].">
<input type= 'submit' name= 'type'
value= 'View/Amend Project Details'></form></td>\n";
}
echo "</table></div>";
I have this on my next page in a table:
$projectid= $_POST['projectid'];
echo "<td>" . $row['projectname'] . "</td>";
I still cannot see the problem, any idea?
The problem is that you're trying to use the $row on your second page, and it isn't set there.
You have to either do the mysqli query again, or transfer the value of $row['projectname'] through the form, using a hidden input field.
<?php
while($row = mysqli_fetch_array($result,,MYSQLI_ASSOC))
{
echo " name='projectid' value=".$row['projectid']."> value= 'View/Amend Project Details'>\n";
}