How to get value from select query without using while loop while we know that output is defiantly only one record
$sql = "SELECT id FROM MyGuests";
$result = $conn->query($sql);
while($row = $result->fetch_assoc())
{
echo $row["id"];
}
Here if i know that there is only one record comes as a output then how to avoid while loop and directly get our id
just delete the while loop!
$sql = "SELECT id FROM MyGuests";
$result = $conn->query($sql);
$row = $result->fetch_assoc();
echo $row["id"];
are you using mysql_* functions by any chance? please switch to PDO as soon as possible.
You can use MYSQl bind result if its a single row output
if( isset($con) && !empty($con) && $con!="" ) {
$knownStmt=mysqli_prepare($con, "SELECT id FROM MyGuests");
if( $knownStmt ) {
mysqli_stmt_execute($knownStmt);
mysqli_stmt_bind_result($knownStmt,$id);
mysqli_stmt_fetch($knownStmt);
mysqli_stmt_close($knownStmt);
}
}
Please try this. This is one of the best way. You can also pass the where condition also and bind the value this query. Please see below is the example for the same.
if( isset($con) && !empty($con) && $con!="" ) {
$knownStmt=mysqli_prepare($con, "SELECT name FROM MyGuests WHERE id=?");
if( $knownStmt ) {
mysqli_stmt_bind_param($knownStmt,"d",$UID);
mysqli_stmt_execute($knownStmt);
mysqli_stmt_bind_result($knownStmt,$Name);
mysqli_stmt_fetch($knownStmt);
mysqli_stmt_close($knownStmt);
}
}
I'm sure this will definitely help you.
Please note this works only for single row result.
You can use this code for your purpose.
$result = mysql_query("SELECT id FROM MyGuests");
$row=mysql_fetch_array($result);
echo $row["id"];
Related
I have a column in a table that I would like to add up and display the result in a textbox. But its not working.
<?php
function getpoint(){
$query = "SELECT sum(monthly_point) FROM characters WHERE
inmate_id = '$value'";
$result_set = mysql_query($query);
$row = mysql_fetch_array($result_set);
echo 'sum:'. $row[0];
}
?>
Switch to mysqli instead of mysql functions (out of date and not as secure) and parameterized statements. Your code is susceptible to SQL injection.
Two problems in your code. Add an alias to the sum column, and reference the column in the echo statement:
<?php
function getpoint() {
$query = "SELECT sum(`monthly_point`) as `sum` FROM characters WHERE inmate_id = '$value'";
$result_set = mysql_query($query);
$row = mysql_fetch_array($result_set);
echo 'sum:'. $row[0]['sum'];
}
?>
When you use var_dump() to display $row then it's OK ?
Please look it and add parameter as below
echo 'sum:'. $row['xxxxx'];
'xxxxx' : parameter
I'm taking data from mysql database table. The table have ID and "POST" columns. I've ordered posts by id's from bottom so i always have the newest post on the first place. But when i want to echo specific post (eg. with id 5) i can't echo it with $col = mysqli_fetch_array($result); echo $col;. I've tried with foreach loop but it echo's all posts. So I thought if i could put them into array with foreach loop it would do the job.
$sql = "SELECT * FROM `post` ORDER BY `id` DESC";
$result = mysqli_query($con, $sql);
$col = mysqli_fetch_array($result);
foreach($col as $cols) {
}
I've tried a lot of things and spent a lot of time on research but still don't have idea how to do it.
Thanks for your ideas and help.
$sql = "SELECT * FROM `post` ORDER BY `id` DESC";
$result = mysqli_query($con, $sql);
$col = mysqli_fetch_array($result);
foreach($col as $cols) {
if($col['id'] == 5) {
print_r($col);
}
}
mysqli_fetch_array fetchs a result row as an associative, a numeric array, or both.
You need to specify the name of the column you want to print out.
Because you may have more than one row in your result set, you should use a loop (while) like so:
while ($row = mysqli_fetch_array($result)) {
echo $row['POST']; // 'POST' here is the name of the column you want to print out
}
Hope this helps!
UPDATED:
If you want to get a specific post, you have to change your SQL to something like this:
$sql = "SELECT * FROM `post` WHERE `id` = $wanted_post_id";
I want to show user if he liked a image or not..
for that I am creating php code
$userid=$_COOKIE['userid'];
$sql = "SELECT * FROM likes WHERE `user_id`='{$userid}'";
$query = mysqli_query($conn, $sql);
$row = mysqli_fetch_assoc($query);
if($row){
echo "unlike";
}
else{
echo "like";
}
I can not do this for everything like 'tags', 'shares', 'comments', 'favourites' ...many
Isn't there anything simpler than this...?
Like say $row_check=mysqli_check_exist($table,$column_name,$userid);
use mysql fetch row method
$num_row = mysqli_num_rows($query);
if($num_row>0)
{
//add your code
}
else
{
//add your code
}
There are a lot of ways of doing this really but if you arnt going to use any more information then weither or not the user has liked it doing select * is a bad idea. The reason why is that you are asking the database to return the value of every column in that table.
Assuming its a small database its probably not a problem no but as your database gets bigger you are puting more load on it then you need you should try and only select the columns you need and intend to use. Ok in this case the userid is probably indexed and its only one row, but if you get in the habit of doing it here you may do it else where as well.
try this instead.
$userid=$_COOKIE['userid'];
$sql = "SELECT count(user_id) as total FROM likes WHERE `user_id`='{$userid}'";
$query = mysqli_query($conn, $sql);
$row = mysqli_fetch_assoc($query);
if( $row ['total'] > 0){
echo "unlike";
}
else{
echo "like";
}
This way we are just getting the total. simple and elegant
Use mysqli_num_rows($query) if > 0 exist
You simply need to count the available records using
mysqli_num_rows($query);
This will return a number (count) of available records
So simple put a check like this :
$userid=$_COOKIE['userid'];
$sql = "SELECT * FROM likes WHERE `user_id`='{$userid}'";
$query = mysqli_query($conn, $sql);
$count = mysqli_num_rows($query);
if($count>0){
echo "unlike";
}
else{
echo "like";
}
I'm working with a file and I'm attempting to do multiple select statements one after another and insert some values. So far the insert and the select I've got working together but when attempting to get the last SELECT to work I get no value. Checking the SQL query in workbench and everything works fine. Here's the code:
$id = "SELECT idaccount FROM `animator`.`account` WHERE email = '$Email'";
$result = mysqli_query($dbc, $id) or die("Error: ".mysqli_error($dbc));
while($row = mysqli_fetch_array($result))
{
echo $row[0];
$insert_into_user = "INSERT INTO `animator`.`user` (idaccount) VALUES ('$row[0]')";
}
$select_userid = "SELECT iduser FROM `animator`.`user` WHERE iduser = '$row[0]'";
$results = mysqli_query($dbc, $select_userid) or die("Error: ".mysqli_error($dbc));
while($rows = mysqli_fetch_array($results))
{
echo $rows[0];
}
I do not want to use $mysqli->multi_query because of previous problems I ran into. Any suggestions? And yes I know the naming conventions are close naming... They will be changed shortly.
Your code makes no sense. You repeatedly build/re-build the $insert_int-User query, and then NEVER actually execute the query. The $select_userid query will use only the LAST retrieved $row[0] value from the first query. Since that last "row" will be a boolean FALSE to signify that no more data is available $row[0] will actually be trying to de-reference that boolean FALSE as an array.
Since you're effectively only doing 2 select queries (or at least trying to), why not re-write as a single two-value joined query?
SELECT iduser, idaccount
FROM account
LEFT JOIN user ON user.iduser=account.idaccount
WHERE email='$Email';
I'm not sure what you're trying to do in your code exactly but that a look at this...
// create select statement to get all accounts where email=$Email from animator.account
$id_query = "SELECT idaccount FROM animator.account WHERE email = '$Email'";
echo $id_query."\n";
// run select statement for email=$mail
$select_results = mysqli_query($dbc, $id_query) or die("Error: ".mysqli_error($dbc));
// if we got some rows back from the database...
if ($select_results!==false)
{
$row_count = 0;
// loop through all results
while($row = mysqli_fetch_array($result))
{
$idaccount = $row[0];
echo "\n\n-- Row #$row_count --------------------------------------------\n";
echo $idaccount."\n";
// create insert statement for this idaccount
$insert_into_user = "INSERT INTO animator.user (idaccount) VALUES ('$idaccount')";
echo $insert_into_user."\n";
// run insert statement for this idaccount
$insert_results = mysqli_query($dbc, $insert_into_user) or die("Error: ".mysqli_error($dbc));
// if our insert statement worked...
if ($insert_results!==false)
{
// Returns the auto generated id used in the last query
$last_inisert_id = mysqli_insert_id($dbc);
echo $last_inisert_id."\n";
}
else
{
echo "insert statement did not work.\n";
}
$row_count++;
}
}
// we didn't get any rows back from the DB for email=$Email
else
{
echo "select query returned no results...? \n";
}
I have the following code and it should return just one value (id) from mysql table. The following code doesnt work. How can I output it without creating arrays and all this stuff, just a simple output of one value.
$query = "SELECT id FROM users_entity WHERE username = 'Admin' ";
$result = map_query($query);
echo $result;
I do something like this:
<?php
$data = mysql_fetch_object($result);
echo $data->foo();
?>
You have to do some form of object creation. There's no real way around that.
You can try:
$query = "SELECT id FROM users_entity WHERE username = 'Admin' ";
//$result = map_query($query);
//echo $result;
$result = mysql_query($query); // run the query and get the result object.
if (!$result) { // check for errors.
echo 'Could not run query: ' . mysql_error();
exit;
}
$row = mysql_fetch_row($result); // get the single row.
echo $row['id']; // display the value.
all you have is a resource, you would still have to make it construct a result array if you want the output.
Check out ADO if you want to write less.
Not sure I exactly understood, what you want, but you could just do
$result = mysql_query('SELECT id FROM table WHERE area = "foo" LIMIT 1');
list($data) = mysql_fetch_assoc($result);
if you wish to execute only one row you can do like this.
$query = "SELECT id FROM users_entity WHERE username = 'Admin' ";
$result = mysql_query($query);
$row = mysql_fetch_row($result);
echo $row[0];
there have been many ways as answered above and this is just my simple example. it will echo the first row that have been executed, you can also use another option like limit clause to do the same result as answered by others above.