I'm trying to populate my dropdown list with informations of my DB. I Use this code:
<select name="teste">
<?php
include "conecta.php";
$sql = sqlsrv_query("select * from carro");
sqlsrv_query($conn, $sql);
while ($row = sqlsrv_fetch_array($sql)) {
echo "<option value=\"teste1\">" . $row['placa'] . "</option>";
}
sqlsrv_close($conn);
?>
</select>
But don't appear any information in my list.
Someone can help me to figure whats wrong in my code?
You are not making the query to the database correctly. This is then causing a PHP error and hence not executing the while loop. The line should be:
$sql = sqlsrv_query($conn, "select * from carro");
or
$query = "select * from carro";
$sql = sqlsrv_query($conn, $query);
You might also want to specify that you are wanting the associative array from the query response, as you are looking for $row['placa']. Like so:
while ($row = sqlsrv_fetch_array($sql, SQLSRV_FETCH_ASSOC)){...}
Related
ok, so I have a search script with is sort of working but I dont get the results I want.
when i run the query in php it doesnt result anything.
$searchStr = htmlspecialchars($searchStr);
$sql = "SELECT * FROM steamitems WHERE steamid='" . $id . "' AND name LIKE '%".$searchStr."%'";
$r_query = mysqli_query($link, $sql);
but if I run the exact same output as $sql (SELECT * FROM steamitems WHERE steamid='76561198196240283' AND name LIKE '%sawed%') in phpmyadmin it returns the correct result..
EDIT: I forgot to mention that I obviously print the results here
while ($row = mysqli_fetch_array($r_query)){
echo $row["assetid"];
echo $row["name];
}
In the below script I want to fetch data from mysql using a explode function and also a variable within an explode function.
Here's how I want to get
<?php
include ('config.php');
$track = "1,2,3";
$i = 1
$trackcount = explode(",",$track);
$sql = "SELECT * FROM tracks WHERE id='$trackcount['$i']'";
$retval = mysql_query($sql, $conn);
while ($row = mysql_fetch_array($retval, MYSQL_ASSOC))
{
echo "{$row['track_name']}";
}
mysql_free_result($retval);
?>
This is the code
$sql = "SELECT * FROM tracks WHERE id='$trackcount[$i]'";
I want sql to fetch data from tracks table where id = $trackcount[$i]
Whatever the value of $trackcount[$i] mysql should fetch but it shows a blank screen.
If I put this
$sql = "SELECT * FROM tracks WHERE id='$trackcount[1]'";
It works perfectly
save your $trackcount[$i] in one variable and then pass it in the query as given below
<?php
include ('config.php');
$track = "1,2,3";
$i = 1;
$trackcount = explode(",",$track);
$id=$trackcount[$i];
$sql = "SELECT * FROM tracks WHERE id='$id'";
$retval = mysql_query($sql, $conn);
while ($row = mysql_fetch_array($retval, MYSQL_ASSOC))
{
echo "{$row['track_name']}";
}
mysql_free_result($retval);
?>
and one more thing check your previous code with echo of your query and see what is passing ok.
echo $sql = "SELECT * FROM tracks WHERE id='$trackcount['$i']'";//like this
problem is with your query
$sql = "SELECT * FROM tracks WHERE id='$trackcount['$i']'";//change
to
$sql = "SELECT * FROM tracks WHERE id='$trackcount[$i]'";
Generally you would want to use the IN operator with this type of query, so for you this would be:-
$sql="SELECT * FROM `tracks` WHERE `id` in (".$track.");";
or, if the $ids are in an array,
$sql="SELECT * FROM `tracks` WHERE `id` in (".implode( ',', $array ) .");";
I have a problem with getting the right value after I counted the rows from a table. I searched on the web but didn't find an answer.
In the database i have a table with all the categories in it they all have an id, and i would like to count using this column.
I have this PHP code, it works but is there an other and better to get over this?
$sql2 = "SELECT COUNT(id) FROM categories";
$stmt2 = sqlsrv_query($conn, $sql2);
$res = sqlsrv_fetch_array($stmt2, SQLSRV_FETCH_ASSOC);
foreach($res as $row)
{
$rows = $row;
}
//if there are categories display them otherwise don't
if ($rows > 0)
{
$sql = "SELECT * FROM categories";
$stmt = sqlsrv_query($conn, $sql);
while ($row = sqlsrv_fetch_array($stmt, SQLSRV_FETCH_ASSOC)) {
echo "<a href='#' class='cat_links'>" . $row['category_name'] . " - <font size='-1'>" . $row['category_description'] . "</font></a>";
}
}
else
{
echo "<p style='text-align: center'>No categories yet.</p>";
}
I think has to be a better way to convert the $stmt2 variable from a SQL resource to an actual number, or to convert the $res variable from an array to an number. If I try to echo the whole array using foreach, it will only print out the number of rows. This is why I use it to count the rows now.
I can't use the sqlsrv_num_rows function because I then get an error, or no answer.
I tried to write my code as short as possible but I discovered something strange here.
If I fetch the query within a 'while loop' the system crashes.
Here's an example.
$sql = 'SELECT * FROM table';
while ($row = $db_ob->query($sql)->fetch_array()){
echo $row['one'];
}
Is this due to my machine or what can I do?
If I write it like my second example, there are no problems
$data = $db_ob->query($sql)->fetch_array()['one'];
This is because you re-run the query on every loop iteration, so it will never end. You need to be sure to only run the query once, then iterate over the results.
$sql = 'SELECT * FROM table';
$result = $db_ob->query($sql);
while ($row = $result->fetch_array()) {
echo $row['one'];
}
Your script execute mysql Query, Retriving datas 'N' times, So you need to rewrite as like,
$sql = 'SELECT * FROM table';
$Result = $db_ob->query($sql);
while ($row = $Result->fetch_array()){
echo $row['one'];
}
Well first of all I am French so I hope my question will be understandable ;-)
I know some people have already experienced problems with queries in php that worked in phpmyadmin. The thing is that each time (or so) these people had "echo" their queries and copy/paste in phpmyadmin, but as php does not always display spaces it was always the problem.
Actually my problem is different :
if I use the query
$sql = "SELECT * FROM jos_dtregister_invoice_sent";
$query = mysql_query($sql);
$result = mysql_fetch_array($query);
it returns all the rows in both phpmyadmin and my php code, but if I want to look in a different table (with same structure), it just works in phpmyadmin and not via my php code (only one row instead of all of them)
Here is the query not working:
$sql = "SELECT * FROM jos_dtregister_receipt_sent";
$query = mysql_query($sql);
$result = mysql_fetch_array($query);
Perhaps the answer is very simple but I admit this is kind of tricky for me...
Many thanks in advance!
Here is my complete code :
function sendReceipt($row) {
$to = getUserInformation($row,10);
$from = getEventAdminEmailFromEmail($row);
$subject = getEventTitle($row)." - Invoice #".$row["confirmNum"];
$message = $row["userFirstName"]." ".$row["userLastName"]." \n\n".getMessageToSendUser($row);
$headers = "From: ".$from."\r\n";
$sql1 = "SELECT * FROM `jos_dtregister_invoice_sent`";
$query1 = mysql_query($sql1);
echo 'Fetched rows number: '.mysql_num_rows($query1)."<br />";
while($row1 = mysql_fetch_array($query1)) {
echo "Invoice Sent: ".$row1["sent"]."<br />";
}
$sql2 = "SELECT * FROM `jos_dtregister_receipt_sent`";
$query2 = mysql_query($sql2);
echo 'Fetched rows number: '.mysql_num_rows($query2)."<br />";
while($row2 = mysql_fetch_array($query2)) {
echo "Receipt Sent: ".$row2["sent"]."<br />";
}
}
mysql_fetch_array() fetches only a row.The name suggests that it fetches a result row as an array
$sql = "SELECT * FROM jos_dtregister_receipt_sent";
$query = mysql_query($sql) or die(mysql_error());
$result = mysql_fetch_array($query);
Try to execute this. The mysql_error() line will spit out what exactly went wrong when trying to execute the SQL. Post it back here and we can help you a little more.
$sql = "SELECT * FROM jos_dtregister_receipt_sent";
$result = mysql_query($sql);
while($row = mysql_fetch_assoc($result)){
// $row will have your data
}
You need to loop over the results with mysql_fetch_array().
while ($result = mysql_fetch_array($query)) {
print_r($result);
}
Or if you want all your results in one big array ($results):
$results = array();
while ($row = mysql_fetch_array($query)) {
$results[] = $row;
}