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";
Related
I come to you in dire need.
What I'm trying to do is to make a form with a dropdown menu, that menu is filled by fetching id's from a table.
<form method="post" action="addWorkExperience.php">
<select name="employerSelection">
<?php
$sql = mysqli_query($conn, "SELECT employer_id FROM employers");
while ($row = $sql->fetch_assoc()){
echo "<option value=\"eID\"> " .$row['employer_id'] . "</option>";
}
?>
</select>
And to my understanding my selected option should be accessible through the $_POST variable like so:
$eValue = $_POST['employerselection'];
or:
$sValue = $_POST['eID'];
These variables would be used in my insert query to write my choice back to the SQL database. However, for some reason it just doesn't. Every other inputfield I trow at it works, except the select field.
Am I missing something?
You have a typo, instead of
$eValue = $_POST['employerselection'];
try
$eValue = $_POST['employerSelection'];
In further situations, you can help yourself by simply using
var_dump($_POST);
that will output all variables that $_POST contains
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 using function to display all values from tables but the ones that are checked in another table to be checked.
I am trying to print all values from one table, print them in checkbox. Then I am using this function to display all values and mark values checked if they are selected in another table.
function emarketing_oprema_checked(){
$id = test_input($_GET['id']);
$link = new mysqli("localhost", "xxx", "xxx", "xxx");
$link->set_charset("utf8");
$sql=mysqli_query($link, "SELECT a.OpremaId, a.OpremaNaziv, b.Oprema, b.Partner
FROM jos_ib_emarketing_oprema a
LEFT JOIN jos_ib_emarketing_stavke_oprema b ON a.OpremaId = b.Oprema where Partner = $id ");
while($record = mysqli_fetch_array($sql)) {
$data='<input type="checkbox" name="oprema[]" value="'.$record["OpremaId"].'"';
if(isset($record['Oprema'])) {//field in the database
$data.=' checked="checked';
}
$data.='">'. $record["OpremaNaziv"] .'</br>';
echo $data;
}
}
As a result I got only checked values, not the ones that are not checked.
Here is my form
<div class="col-xs-6">
<input type="checkbox" id="oprema" onclick="Exposeoprema()">Oprema<br>
<div id="Scrolloprema" style="height:150;width:200px;overflow:auto;border:1px solid blue;display:none">
<?php
emarketing_oprema_checked();
?>
</div>
</div>
The isset($record['Oprema'] will always be true, even if the value of it is null because the field is exists.
You need to check the value of it.
Try:
if ($record['Oprema']) {
$data.=' checked="checked';
}
In insert form I am using check boxes to insert values in database. Now, I want to get checked values in edit form so I can check new values or uncheck checked values.
First I get values from database (I am using pdo, I will not post connection code to db- it works):
get oprmea from database
$sql_oprema = "SELECT Oprema FROM dbo_emarketing_stavke_oprema WHERE Partner='$id'";
$n = $conn->query($sql_oprema);
while ($r = $n -> fetch()) {
$oprema_sql = $r['Oprema'];
}
I get values when I dump variables, output is not NULL.
I am using function to store values in database. Now I want to use same function for editing if posssible.
function emarketing_oprema(){
$link = new mysqli("localhost", "root", "le30mu09", "websoft");
$link->set_charset("utf8");
$sql=mysqli_query($link, "SELECT * FROM `dbo_emarketing_oprema` order by OpremaId asc ");
while($record = mysqli_fetch_array($sql)) {
echo '<input type="checkbox" name="oprema[]" value="'.$record['OpremaId']. '">' . $record['OpremaNaziv'] . ' <br/><br/> </input>';
}
}
I was wondering is it possible to use this function to get checked values checked.
use checked attribute
<input type="checkbox" checked>
or like this
<input type="checkbox" checked="checked">
You php will look like this:
while($record = mysqli_fetch_array($sql)) {
data='<input type="checkbox" name="oprema[]" value="'.$record["OpremaId"];
if(isset($record['checked'])) {//field in the database
data+=' checked="checked';
}
data+='">'. $record["OpremaNaziv"].'</br>';
}
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.