I am trying to create a form using data from MySQL and my current output is not looking as expected so I just need some simple changes to my current form
My current output in the browser: http://gyazo.com/f4668ca59586ec0d4d1500ea6f7b6257
What I am looking to do is
fix the problem with the first column called 'Long Jump A' so its
next to a drop down menu, and you can see the bottom one is next to
nothing.
Secondly, How can I swap the order of the drop down and the
event? So the event is on the left then the drop down is on the
right?
Moving the form down from the top so there is a space
The code:
<?php
require_once 'db/connect.php';
//Query to display all events
if ($event_result = $con->query("SELECT Event.Name FROM event")) {
echo "<form method =\"POST\">";
while ($row = $event_result->fetch_assoc()) {
echo $row['Name'] . ' <br> ';
if ($student_result = $con->query("SELECT Student.Form, Teacher.Form, Student.Forename, Student.Surname, Student_ID " .
"FROM Student, Teacher " .
"WHERE Student.Form = Teacher.Form AND Teacher.Form = 'C'")) {
if ($student_result->num_rows) {
echo "<select name ='Student_ID'>";
while ($row1 = $student_result->fetch_assoc()) {
echo "<option value ='" . $row1['Student_ID'] . "'>" . $row1['Forename'] . ' ' . $row1['Surname'] . "</option>";
}
echo "</select>";
}
}
}
echo "</form>";
}
?>
You have an opening <br /> as you echo out the name of the row. This means echo the title, then break before you display the <select> group. You should move that <br /> down right before the closing while loop.
Or just remove the top <Br /> and add it back after the ending so it looks like this:
echo "</select> <br />";
edit: this will fix the last two of your questions, as now each select group will properly line up with the title of the row. If you want more space at the top of the form you have a bunch of options: add more breaks, add a margin or padding, or position the element as relative and use css top property or CSS transforms to move it.
Related
When I create/edit post to my website, and when I use html tags (either type it by my self or using WYSIWYG editor (in this case nicEdit, but I've used different one - same problem)) it does not show anything on the page.
to be more specific: I have article container which imports php code, which than iterates through the db table and echo posts. This is the code:
<h3>Wydarzenia</h3>
<?php
//collects data from table "events" and sort it by "id" date in descending oreder
$event_table_data = mysqli_query($db_con, "SELECT * FROM events ORDER BY id DESC")
or die(mysqli_error());
//puts the data from "events" into an array
$event_content_array = mysqli_fetch_array($event_table_data);
//makes variables from table columns
$id = id;
$title = title;
$happeninng_date = happen;
$created_date = created;
$content = content;
$author = author;
//add new post button
if((isset($_SESSION['sess_wizardrights'])) && ($_SESSION['sess_wizardrights'] == true)){
echo "<a class='e_event_addpost_link' href='./index.php?link=add_post'><button class='e_event_addpost_button' type='button'>Dodaj nowy post</button></a>";
}
do{
//echo each event for main event article
echo "<span class='b_events_span' id='" . $event_content_array[$id] . "'>";
echo " <a class='b_events_anchor' id='" . $event_content_array[$id] . "'></a>";
echo " <h2 class='b_events_title'>" . $event_content_array[$title] . "</h2>";
if((isset($_SESSION['sess_wizardrights'])) && ($_SESSION['sess_wizardrights'] == true)){
echo "<form class='b_events_edit_form' name='edit_post' action='./index.php?link=edit_post' method='post'>";
echo " <input type='hidden' name='id' value='" . $event_content_array[$id] . "'> ";
echo " <input class='e_events_edit_submit' type='submit' value='Edytuj wydarzenie'>";
echo "</form>";
}
echo " <div class='fb-share-button' data-href='http://www.zhppgkaberdeen.co.uk/index.php?link=events#" . $event_content_array[$id] . "'></div>";
echo " <p class='b_events_happening_date'><i>Data wydarzenia: " . $event_content_array[$happeninng_date] . "</i></p>";
echo " <p class='b_events_content'>" . $event_content_array[$content] . "</p>";
echo " <p class='b_events_created_date'>Utworzono: " . $event_content_array[$created_date] . "</p>";
echo " <p class='b_events_author'><b>Czuwaj!</b><br/>" . $event_content_array[$author] . "</p>";
echo "</span>";
}while($event_content_array = mysqli_fetch_array( $event_table_data ));
?>
When I add tags like <img> for instance with alt="" parameter it completely ruins web page. What I mean by that is that the article container is not shown at all (I've checked that with firebug) as well as everything after the article tag (like footer for instance). What is even more strange is that this: alt="text" did not break the page but every other random word does, how ever it does not show the "text" at image at all.
Similar thing happen when I tried to create table within the post and title or target parameter, I cant remember which one.
I have tried to use htmlentities($event_content_array[$content]), htmlspecialchar($event_content_array[$content]) and mysqli_real_escape_string($do_con, $event_content_array[$content]) -> non of this helped.
Is there a way to fix that or I just need to give up using this tags/parameters?
Thank you for any help or explanation
if you are using e.g echo "<img alt="text">";try to escape the " such as alt=\"text\"
First off, I want to store the names of these checkboxes which are submitted, and not their values.
This is my code:
<?php
$con=mysqli_connect("localhost","root","","notifier");
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con,"SELECT * FROM student");
echo "Enter the attendance. Please untick for 'ABSENT' students and submit";
echo "<br>";
echo "<form action=\"d.php\" method=\"post\">";
while($row = mysqli_fetch_array($result))
{
echo "<br>" .$row['classrollno'] . "   <input type=\"checkbox\" name=\"" . $row['studentid'] . "\" value=\"P\" checked>";
}
echo "<input type=\"submit\" name=\"submit\" value=\"submit\">";
echo "</form>";
?>
This code simply fetches a column of student rollnumberss from student table, prints them, and as well as prints a checkbox infront of them which is checked by default.
Names of checkboxes will be the student id (varchar, another column).
Now since All Checked checkboxes, that is the checboxes which will be submitted to next page will have same default value "P", I m not concerned about their values.
How do I store the names of these checkboxes in an array, and later on use it to perform updation in table for all these student id's?
Use the following code:
while($row = mysqli_fetch_array($result))
{
echo '<br>' .$row['classrollno'] . ' <input type="checkbox" name="studentId[]" value="' . $row['studentid'] . '" checked />';
}
Then, when you process the form, the $_POST['studentId'] variable will contain an array with all the id's.
Since the value that will probably be inserted in the db is 'P' for every student, you wouldn't need to include it in your form, but just hardcode it in your query.
Keep adding the names to an array. Its straight forward.
Declare $allStudentIds = array(); outside while loop. Then, to store in that array,
$allStudentIds[] = $row['studentid'];
Since you wanted to use these values later, you can directly store them inside a session variable:
$_SESSION['allStudentIds'][] = $row['studentid'];
In above case, $_SESSION['allStudentIds'] will be an array of all student ids selected.
Note: You need to start session using session_start() as the first line in the script after opening <?php tag.
Simply, in the fetching while loop, define an array and set each checkbox value to one of its elements then assign it as a session variable:
while($row = mysqli_fetch_array($result))
{
echo "<br>" .$row['classrollno'] . "   <input type=\"checkbox\" name=\"" . $row['studentid'] . "\" value=\"P\" checked>";
$names[] = $row['studentid'];
}
Then,
$_SESSION['names'] = $names;
Your confusion seems to stem from the fact that you are mixing the View (the name of the checkbox in HTML) and the Model/Data (which the student_id you are getting from your DB query ie. the $row = mysqli_fetch_array($result) in the while loop).
All you need to do is create an empty array (eg. $studentid_arr) before the loop and after the echo statement which is just contributing to the view (the HTML) you do some work with your data. What you want to do currently is to store the student_ids (and not the name of the checkbox) in your $studentid_arr.
That can be done with a simple array_push ($studentid_arr,$row['studentid']);
So your while loop would look like
while($row = mysqli_fetch_array($result))
{
echo "<br>" .$row['classrollno'] . "   <input type=\"checkbox\" name=\"" . $row['studentid'] . "\" value=\"P\" checked>";
array_push ($studentid_arr,$row['studentid']);
}
Now you can just POST this PHP array to your next script which is expecting these values. (which is what I assume you mean by submitting to the next page)
First of all, I am a newbie when it comes to coding, so please be kind and patient :)
What I am trying to do is to select two rows ('ID', 'name') from a MySQL table (categories), populate a drop down list with one row ('name'), and on submission of a form, pass the other ('ID') to another table.
Now, I can populate the drop down list, no problem. I have populated this with both 'ID' and 'name' to test that both of the variables I am using to hold this information, contain the correct data. But I cannot seem to $_POST the information.
I guess I am either looking at the wrong part of the array, or I am simply using the wrong code.
This is the code to create a new product, under a category from the database.
<?php
include 'db_config.php';
?>
<form enctype="multipart/form-data" action="insert.php" method="post">
<h3>Add New Product</h3>
Category:
<!-- START OF categories (table) names (row) SQL QUERY -->
<? $sql = "SELECT ID, name FROM categories";
$result = $mysqli->query($sql);
echo "<select name='category_name'>";
while ($row = $result->fetch_assoc()) {
$cat_ID=$row['ID'];
$cat_name=$row['name'];
extract($row);
echo "<option value='" . $cat_ID . $cat_name . "'>" . $cat_ID . " " . $cat_name ."</option>";
}
echo "</select>";
?>
<!--END OF SQL QUERY -->
<br>
Code: <input type="text" name="code"><br>
Name: <input type="text" name="prod_name"><br>
Description: <input type="textarea" name="description"><br>
Image: <input type="file" name="image"><br>
<input type="Submit">
</form>
For now, I am just echoing this out in the insert.php script, to test the code above. This is a snippet of the insert.php script.
echo "ID: " . $_POST['$row["ID"]'] . "<br>";
echo "Category: " . $_POST['$row["name"]'] . "<br>";
echo "Code: ". $_POST['code'] . "<br>";
echo "Name: " . $_POST['prod_name'] . "<br>";
echo "Description: ". $_POST['description'] . "<br>";
echo "Image: " . $_POST['image'] . "<br>";
Don't worry about the last line above. I know this needs to be $_FILES, and I have all this covered. I have stopped writing the data to the table until I get my issue fixed. In the full script, image are being upload to "/images" and the location stored in the table. This all works fine.
The problem is with the first two lines, as they are blank when returned. I thought I was storing the information correctly, as I am calling the same variables to populate the drop down list, but I cannot seem to $_POST it.
Does that makes sense?
Thanks to all who help me. Once day I will be as good as you....I hope.
TIA
Smurf.
this bellow:
echo "ID: " . $_POST['$row["ID"]'] . "<br>";
echo "Category: " . $_POST['$row["name"]'] . "<br>";
is wrong, select element has its name category_name, so, instead of this, you should
do:
echo "Category: " . $_POST['category_name'] . "<br>";
echo "ID: " . $_POST['$row["ID"]'] . "<br>";
echo "Category: " . $_POST['$row["name"]'] . "<br>";
There aren't any form elements with those names in your form ($row["ID"] and $row["name"]). Those would be really strange names for a form element anyway. The form element you're creating is:
<select name='category_name'>
So the selected value would be posted as:
$_POST['category_name']
The option elements for that select appear to have values which are a combination of ID and Name:
"<option value='" . $cat_ID . $cat_name . "'>"
Thus, if the user selects an option with a value of 1SomeName then $_POST['category_name'] will evaluate to '1SomeName'.
It's certainly unconventional to use the combination of ID and Name for the option values, but it should work. The problem is presents is that you now have a composite string which needs to be parsed in order to be useful. Generally what one would do is just use the ID as the value and the Name as the display. All you should need to use it throughout the code is the ID.
The $_POST variable you want, is inside category_name
Cuz your select is...
<select name='category_name'>
So you need to get it by...
$_POST['category_name'];
Which will return whatever you've assigned to the select options...ie
2 Name
2 being the ID, and Name being the name
But if you then want to use that ID to retrieve from DB or anything, you're gonna have to explode that apart...like so....to get each piece.
$array = explode(" ", $_POST['category_name']);
That will leave you with...
$array[0] = ID
$array[1] = Name
But I would avoid all that part, by just assigning the ID to the value only...like so..
echo "<option value = '".$cat_ID."'> ".$cat_name." </option>";
That way you just pass the ID and have access to it on the other side.
I have a database that I want to get data out onto a website. It contains states listed by name and id. Counties listed by id, namne , and state that contains thems ID and then clubs that exist , with a reference to the county id's that they exist in and columns for their actual data.
What I've got :
A drop down menu that populates itself with state id and name.
What I'd like to accomplish:
On selection of state , let's say ny , take it's id and use this in gathering another mysql array for the county drop down. I'd like it to dynamically occur on selection of state , maybe even giving a count of results next to the drop down.
$resstate = mysql_query("SELECT * FROM state ORDER by longstate;") or die("Note: " . mysql_error());
State:
<select name="State" size=1>
<?
while( $rs = mysql_fetch_array( $resstate ) ) {
echo "<option value=" .$rs['id'] . ">" . $rs['longstate'] . "</option>";
}
echo "</select>";
?>
I know I could use a JavaScript onChange="this.form.submit()" on the first drop down, but it's my understanding that I'd then be making a new page at that point and don't know if I could keep the functionality of the state drop down, say if you accidentally chose new Hampshire when you wanted New York.
here's an example of the current array filling the drop down :
http://snowmobileamerica.com/countytest.php
----EDIT---
Using Dagons Advice , I looked into Ajax.
I made a php file that's supposed to query the database based on a reference to getcounty.php?q=
The file is created as follows :
<?php
$q=$_GET["q"];
$cn=mysql_connect("localhost","user","password") or die("Note: " . mysql_error());
mysql_select_db("snowusa_clubs", $cn);
$sql="SELECT * FROM county WHERE state_id = '".$q."' ORDER by name";
$result = mysql_query($sql);
echo "<select name="County" size=1>";
while($rc = mysql_fetch_array($result))
{
echo "<option value=" .$rc['id'] . ">" . $rc['name'] . "</option>";
}
echo "</select>";
mysql_close($cn);
?>
If i try to run it manually http://www.snowmobileamerica.com/getcounty.php?q=33 I get a 500 internal server error...
Any ideas where I went wrong?
try adding an id to the element, then make an ajax call to a handler with jquery:
$("#State").change(function() {
$.post("path/to/request handler/" , { "State" : $(this).val() },
function(data){
if (data == "OK"){
//add some elements here
} else {
//handle an error here
}
});
});
not able to comment yet.
but for the second question try:
<?php
$q=$_GET["q"];
$cn=mysql_connect("localhost","user","password") or die("Note: " . mysql_error());
echo "Conn ok<br>";
mysql_select_db("snowusa_clubs", $cn);
echo " Database opened<br>";
$sql="SELECT * FROM county WHERE state_id = '$q' ORDER by name";
$result = mysql_query($sql);
echo " Database queried <br>";
echo "<select name='County' size=1>";
while($rc = mysql_fetch_array($result))
{
echo "<option value='" .$rc['id'] . "'>" . $rc['name'] . "</option>";//added single quotes in the value
}
echo "</select> ";
mysql_close($cn);
?>
I have a PHP form, which in the end is going to write all the filled in data to a MySQL database. That's all fairly doable, but here is something tricky that I don't know how to handle:
I have a DIV which requests all the rows of one table. Every row represents a category and the user can choose several categories to add to their post. This is the code I use to retrieve the rows + checkbox in front of them:
<?php
$sql = "SELECT merknaam FROM merken";
$result = mysql_query($sql);
while ($row = mysql_fetch_array($result)) {
echo " <input type=\"checkbox\" name=\"merken\" value='" . $row['merknaam'] . "'> " . $row['merknaam'] . " <Br /> ";
}
?>
The user can choose multiple categories, and I would like to save them all and write them to my database. All the categories should be written to 'one' column called 'categories', but the seperate categories should still be distinguishable.
Might be a tricky one? Hope someone can help!
Change the code for the checkboxes to:
echo " <input type=\"checkbox\" name=\"merken[]\" value='" . $row['merknaam'] . "'> " . $row['merknaam'] . " <br /> ";
Adding the [] after the name will allow PHP to treat the checkboxes as an array, which you can serialize and store in your database.