I have four drop-down lists that I would like to populate with values from an MSSQL table. All four lists should contain the same values. The query looks like this:
$data = $con->prepare("SELECT ID, Code FROM Table WHERE Code = :value ORDER BY Code");
$input = array('value'=>'value'); //'value' is hardcoded, not a variable
$data->execute($input);
And here is the code for my drop-downs:
<?php
echo "<select name=\"proj1[]\">";
while($row = $data->fetch(PDO::FETCH_BOTH))
{
echo "<option value='".$row['Code']."'>".$row['Code']."</option> ";
}
echo "</select>";
?>
This works fine for one drop-down. If I try to create another one (proj2[], proj3[], proj4[]) and apply the same query, however, the PHP page stops loading at that point and the second drop-down does not populate. The only way I've found around it is to copy the query and change the variables ($data becomes $data2 for proj2[], and so on). I'd really rather not have to write the same query four times. Is there a way around it?
$select = '';
while($row = $data->fetch(PDO::FETCH_BOTH))
{
$select .= "<option value='".$row['Code']."'>".$row['Code']."</option> ";
}
echo "<select name=\"proj1[]\">";
echo $select;
echo "</select>";
echo "<select name=\"proj2[]\">";
echo $select;
echo "</select>";
//etc...
Why not just put all of it in a veriable and then using it 4 times?
Somthing like this:
<?php
while($row = $data->fetch(PDO::FETCH_BOTH))
{
$options .= "<option value='".$row['Code']."'>".$row['Code']."</option> ";
}
for($i = 0; $i <= 4; $i++){
echo "<select name=\"proj1[]\">";
echo $options;
echo "</select>";
}
?>
Related
In this code i am trying to fetch city names into the html dropdown. Kindly corrct me if i am wrong anywhere, it give me an error
<?php
$query = Run("select city_name from City");
echo "<select name="city-name" style="width: 210px;">";
while ($row = mssql_num_rows($query))
{
echo "<option>$row->city_name</option>";
}
echo "</select>";
?>
use mysqli_fetch_array($query) insted of mssql_num_rows($query)
try this one
echo "<select>";
while ($row =mysqli_fetch_array($query))
{
echo "<option value='".$row['city_name']."'>".$row['city_name']."</option>";
}
echo "</select>";
mssql_num_rows returns the number of rows in the result set, it doesn't iterate and return individual rows. Try using mssql_fetch_object instead.
If you start a string with " then you have to escape all occurrences of " inside your string or use '.
For example:
echo "<select name=\"city-name\" style=\"width: 210px;\">";
or
echo '<select name="city-name" style="width: 210px;">';
so you don't accidently close the string.
Also like the others pointed out, you have to use
mysqli_fetch_array($query).
Use this
$query = Run("select city_name from City");
echo "<select name='city-name' style='width: 210px;'>";
while ($row = mssql_fetch_object($query))
{
echo "<option>$row->city_name</option>";
}
echo "</select>";
Use This
$query = Run("select `city_name` from City");
echo "<select name='city-name' style='width: 210px;'>";
while ($row = mssql_num_rows($query))
{
echo "<option value='.$row->city_name.'>".$row->city_name."</option>";
}
echo "</select>";
Ok so here is my code:
$select_status = 0;
$select_status = "<select name='status'>\n";
$select_status .= "<option value=''>SELECT ONE</option>\n";
$sdataset = mysql_query("SELECT id, name FROM phponly_category") or die(mysql_error());
while($srow=mysql_fetch_assoc($sdataset)) {
echo implode(", ", $srow);
echo "<br />";
$select_status .= "<option value='".$srow['name']."'";
$select_status .= ">".$srow['name']."</option>\n";
} // end while loop
echo "out of the loop";
$select_status .= "</select>\n";
// now insert the <select> list control into the page
echo $select_status;
The code works fine until the last row when it breaks. I cannot get the echo $select_status printed.
I have tried to see what is going on with the SQL query results by printing each row but everything looks fine there. For some reason, at the last row the while loop breaks and even the code after while loop doesn't get executed.
Don't do the or die(mysql_error()) portion in the while test...do it before.
if($sdataset==false) {
die(mysql_error());
}
while($srow=mysql_fetch_array($sdataset)) {
$select_status .= "<option value='".$srow['name']."'".">".$srow['name']."</option>\n";
} // end while loop
The or die() on your while() loop will actually kill the script when you read the end of the result set. mysql_fetch will return false, triggering the or die().
While checking for errors is good on queries, you can't do it like this on the fetching part, because you get false positives like this.
I personally don't like to echo out html code, if your goal is to do validate whether there's result comes out of the query, you can do something like this
<?php
$sdataset = mysql_query("SELECT id, name FROM phponly_category");
if (mysql_num_rows($sdataset) > 0) {
?>
<select name='status'>
<option value=''>SELECT ONE</option>
<?php
while($srow = mysql_fetch_array($sdataset)) {
?>
<option value='<?php echo $srow['name'] ?>'><?php echo $srow['name'] ?></option>
<?php } // end while loop ?>
</select>
<?php
} // end of if
else {
// Whatever you wanna put here
}
?>
EDITED: There's a typo at mysql_num_rows, try this one again
Use mysql_fetch_assoc so you can get the data using the field name as the key.
<?php
$conn = mysql_connect('localhost', 'username', 'password');
mysql_select_db('test');
$select_status = "<select name='status'>\n";
$select_status .= "<option value=''>SELECT ONE</option>\n";
$sdataset = mysql_query("SELECT id, name FROM phponly_category") or die(mysql_error());
while($srow=mysql_fetch_assoc($sdataset)) {
$select_status .= "<option value='".$srow['name']."'".">".$srow['name']."</option>\n";
} // end while loop
$select_status .= "</select>\n";
echo $select_status;
?>
Note: The use of this extension is discouraged. Take a look at mysqli or PDO.
.hey guys how can i populate a dropdown list with the list of tables from a certain database?
$db = mysql_select_db('thepillar');
$sql = "SHOW TABLES";
$result = mysql_query($sql);
echo '<form method="post" id="try" action="pillar.php">';
echo 'Select Batch: ';
echo '<select name="batch" id="batch">';
echo '<option>';
while($r = mysql_fetch_assoc($result))
{
$tables = $r;
echo '<option>'.$tables.'</option>';
}
.i have tried the code above but the dropdown list is only filled with the word "Array" multiple times depending on how many tables are there in the database.
.help pls!
while($r = mysql_fetch_array($result))
{
echo $r[0]."<br />";
}
replace
$tables = $r;
with
$tables = $r['Tables_in_thepillar'];
also you got an extra echo '<option>';
above the loop
Your $tables variable is an associative array. You need to specify which index of the array you want to output between the <option> tags.
echo '<option>'.$tables['TABLE_NAME'].'</option>';
See the print_r output for what the index name is.
$sql = "SELECT DISTINCT Branch FROM student_main";
$result = mysql_query($sql);
$row_num = mysql_num_rows($result);
$rows = mysql_fetch_array($result);
echo "<select name='Branch'>";
for($i=0;$i<=$row_num-1;$i++){
echo "<option value='".$rows[$i]."'>".$rows[$i]."</option>";
}
echo "</select>";
echo "<input type='submit' Value='submit' />";
echo "</form>";
I am trying to create a dropdown using the above code for my form. But its not working. There are 3 distinct values in the Branch column but in the dropdown, it shows only one value(the first one) and the next two as blank values.
However when in echo $row_num, its shows 3.
Thats means its fetching the three rows, but then why its not showing in the dropdown list.
If I run the same query in phpmyadmin it shows the correct answer i.r it returns 3 distinct Branch values.
You should do something like this:
$sql = "SELECT DISTINCT Branch FROM student_main";
$result = mysql_query($sql);
echo "<select name='Branch'>";
while ($row = mysql_fetch_array($result)) {
echo "<option value='".$row[0]."'>".$row[0]."</option>";
}
echo "</select>";
echo "<input type='submit' Value='submit' />";
echo "</form>";
you need to mysql_fetch_array() for each row. That function returns an associative array for one row only. just include it inside your for loop just above your echo statement.
edit: mysql_fetch_array() actually returns an array (by default) that has associative indices and numbered indices. You can continue using it the same way, though.
You need to loop through your query using the following:
$sql = "SELECT DISTINCT Branch FROM student_main";
$result = mysql_query($sql);
echo "<select name='Branch'>";
while($rows = mysql_fetch_array($result)){ // should probably use mysql_fetch_assoc()
echo "<option value='".$rows['Branch']."'>".$rows['Branch']."</option>";
}
echo "</select>";
echo "<input type='submit' Value='submit' />";
echo "</form>";
mysql_fetch_array only returns the current dataset as an array, and moves the internal pointer ahead. You need to repeatedly call mysql_fetch_array to get all results.
while ($row = mysql_fetch_array($result)) {
echo "<option value='".$row['Branch']."'>".$row['Branch']."</option>";
}
There is a problem in the loop using a while loop:
while($rows=mysql_fetch_array($result)){
echo "<option value='".$rows[$i]."'>".$rows[$i]."</option>";
}
Try this
What you really need is to learn how to use templates.
But it seems Stackoverflow is definitely not the place where one can learn professional ways of website developing.
get your data first
$select = $array();
$sql = "SELECT DISTINCT Branch FROM student_main";
$res = mysql_query($sql) or trigger_error(mysql_error().$sql);
while($row = mysql_fetch_array($res)) $select = $row[];
And then use it in the template
<form>
<select name='Branch'>
<? foreach($select as $row): ?>
<option value="<?=htmlspecialchars($row['Branch'])?>">
<?=htmlspecialchars($row['Branch'])?>
</option>
<? endforeach ?>
</select>
<input type='submit' Value='submit' />
</form>
mysql_fetch_array will only return the first row...
see here for full details :)
I am trying to loop though my users database to show each username in the table in their own row. I can only get it to show one user but loops through this the number of rows there are. Code is below
<?php
require_once ('../login/connection.php');
include ('functions.php');
$query = "SELECT * FROM users";
$results=mysql_query($query);
$row_count=mysql_num_rows($results);
$row_users = mysql_fetch_array($results);
echo "<table>";
for ($i=0; $i<$row_count; $i++)
{
echo "<table><tr><td>".($row_users['email'])."</td></tr>";
}
echo "</table>";
?>
Thanks
mysql_fetch_array fetches a single row - you typically use it in a while loop to eat all the rows in the result set, e.g.
echo "<table>";
while ($row_users = mysql_fetch_array($results)) {
//output a row here
echo "<tr><td>".($row_users['email'])."</td></tr>";
}
echo "</table>";
You're only fetching one row:
$row_users = mysql_fetch_array($results);
You're never calling a fetch again so you're not really looping through anything.
I'd suggest changing your loop to the following:
echo "<table>";
while ($row = mysql_fetch_array($results)) {
echo "<tr><td>".($row['email'])."</td></tr>";
}
echo "</table>";
The while loop will loop through the results and assign a row to $row, until you run out of rows. Plus no need to deal with getting the count of results at that point. This is the "usual" way to loop through results from a DB in php.
In the new MYSQLI, I use this coding
$query = "SELECT * FROM users";
$results=mysqli_query($con,$query);
$row_count=mysqli_num_rows($results);
echo "<table>";
while ($row = mysqli_fetch_array($results)) {
echo "<tr><td>".($row['id'])."</td></tr>";
}
echo "</table>";
mysqli_query($con,$query);
mysqli_close($con);
Your problem is in this line:
echo "<table><tr><td>".($row_users['email'])."</td></tr>";
You're echoing out a <table> tag again. Remove that so your code looks like this:
echo "<table>";
for ($i=0; $i<$row_count; $i++)
{
echo "<tr><td>".($row_users['email'])."</td></tr>";
}
echo "</table>";
You don't need to output a table within a table the way you're doing it.
$result = mysql_query("SELECT `email` FROM `users`");
$num_emails = mysql_num_rows($result);
echo "<table><caption>There are/is $num_emails email(s)</caption>";
while ($row = mysql_fetch_assoc($result)) {
echo "<tr><td>{$row['email']}</td></tr>";
}
echo '</table>';
Something like that should work.