What's the best way to check whether the value is in the database?
Am I doing it correct?
$result = mysql_query("SELECT COUNT(*) FROM table WHERE name = 'John'");
$count = count($result);
you could use straight forward ,
mysql_num_rows() ;
eg :
$con = mysql_connect($host,$uname,$passwd)
mysql_select_db($dbase,$con);
$result = mysql_query($query,$con);// query : SELECT * FROM table WHERE name='jhon';
if( ! mysql_num_rows($result)) {
echo " Sorry no such value ";
}
Yes you are doing it right, if you are only concerned with checking if there are any records where name='john'
SELECT COUNT(*) FROM table WHERE name = 'John'
will return the no. of records where name field is 'John'. if there are no records then it will return 0, and if there are any records it will return the number of records.
But the above query will miss the entries where name is 'John Abraham' or 'V john', to include even these
you can modify your query like this.
SELECT COUNT(*) FROM table WHERE name like '%John%'
I'd say yes.
$result = mysql_query("SELECT COUNT(*) AS 'nb' FROM table WHERE name = 'John'");
$line = mysql_fetch_array($result, MYSQL_ASSOC);
$count = $line['nb'];
Will give you the number of matching rows.
$result = mysql_query("SELECT COUNT(*) as user FROM table WHERE name = 'John'");
$line = mysql_fetch_array($result, MYSQL_ASSOC);
$count = $line['user'];
if($count!=0)
{
echo "user exists";
}
else
{
echo "There is no such user";
}
Related
I want to get all emp_ids of employees which is manager_id equal to login user_id. But why this query show only first result of the column. I could not able to find the problem.
$userID=$_SESSION['userID'];
var_dump($userID);
$result1 = mysql_query("SELECT emp_id FROM employee where manager_id='$userID' ORDER BY emp_id");
$array = mysql_fetch_array($result1);
$id1=$array['emp_id'];
Please help !
You should loop over the result
and eg: you can store all the empID in an array
$cnt = 0;
while ($row = mysql_fetch_array($result1)) {
echo "ID: " . $row[0] ;
$myArrayOfEmpID[$cnt] = $row[0];
$cnt++;
}
instead you only set
$array = mysql_fetch_array($result1);
$id1=$array['emp_id'];
I'm trying to sum my MySQL column value with number, here is my php script, i dont think im doing in right way. I was searching in stackoweflow but i havent found how to sum with specific number, not with other column.
<?php
$currentUser = isset($_POST['currentUser']) ? $_POST['currentUser'] : '';
$currentTasken = isset($_POST['currentTasken']) ? $_POST['currentTasken'] : '';
$con = mysql_connect("localhost", "root", "") or die(mysql_error());
if(!$con)
die('Could not connectzzz: ' . mysql_error());
mysql_select_db("foxi" , $con) or die ("could not load the database" . mysql_error());
$check = mysql_query("SELECT * FROM dotp_task_log");
$numrows = mysql_num_rows($check);
if($numrows >= 1)
{
//$pass = md5($pass);
$ins = mysql_query("INSERT INTO dotp_task_log (task_log_creator, task_log_Task) VALUES ('$currentUser' , '$currentTasken')" ) ;
if($ins)
$check = mysql_query("SELECT * FROM dotp_task WHERE task_id='$currentTasken'");
$numrows = mysql_num_rows($check);
if($numrows == 1)
{
//$pass = md5($pass);
$ins = mysql_query("SELECT (SELECT SUM(task_percent_complete) FROM dotp_task WHERE task_id='$currentTasken') FirstSum, (SELECT SUM(5)), SecondSum ");
if($ins)
die("Succesfully summed!");
else
die("ERROR");
}
else
{
die("Cant sum!");
}
die("Succesfully Created Log!");
else
die("ERROR");
}
else
{
die("Log already exists!");
}
?>
Your query is not written properly. Try to run this query from MySQL command line or from PHPMyAdmin. It will give you an idea what the error is. My best guess is that the first part of the query is not complete.
This is what it will resolve to
SELECT n FirstSum, 5, SecondSum
Where n is the value returned from the subquery. This is a syntax error. Try to remove the last comma before SecondSum.
This is what your query should be:
SELECT (SELECT Sum(task_percent_complete)
FROM dotp_task
WHERE task_id = '$currentTasken') FirstSum,
(SELECT Sum(5)) SecondSum
remove
(select sum(5))
and replace it with
(select 5)
I am very confused about this (returning false):
$sql = "SELECT * from tbl_user WHERE group = 'abc'";
$res = mysql_query($sql);
if(mysql_num_rows($res) > 0) {
$response = array('status' => '1');
} else {
$response = array('status' => '0'); // ---> what I get back
die("Query failed");
}
...despite the fact the field group is present in mySQL database. Even more strange is that the following return the value of group:
$SQL = "SELECT * FROM tbl_user";
$result = mysql_query($SQL);
while ($db_field = mysql_fetch_assoc($result)) {
print $db_field['group']; // ---> returns 'abc'
When I execute a WHERE clause with every other fields of my table excepting group (for example WHERE name = 'ex1' AND ID=1 AND isAllowed=0 (and so on...), everything is fine. As soon as I insert group = 'abc', I get nothing...
This makes me mad. If anyone could help... (I am running a local server with MAMP).
Thanks a lot!
The issue is that group is a reserved word in SQL.
For MySql you need to escape it with backticks
`group`
So your query would be
$sql = "SELECT * from tbl_user WHERE `group` = 'abc'";
I'm a MYSQL/PHP newbie and I'm sure this is a simple question. I'm trying to calculate the average of several questions and respondents from one table and updating a Group table with that value.
For example Table answers consists of (name, group_id, TaskClarity1, TaskClarity2, TaskClarity3) in Table B i want (group_id, avg(TaskClarity1,TaskClarity2,TaskClarity3)).
This is what I've got...
$avg_task_clarity_1 = mysql_query("SELECT AVG(TaskClarity1) WHERE gruppid = '$group_id'");
$avg_task_clarity_2 = mysql_query("SELECT AVG(TaskClarity2) WHERE gruppid = '$group_id'");
$avg_task_clarity_3 = mysql_query("SELECT AVG(TaskClarity3) WHERE gruppid = '$group_id'");
$avg_task_clarity = ($avg_task_clarity_1+$avg_task_clarity_2+$avg_task_clarity_3)/3;
$print_task_clarity_1" UPDATE results SET results.TaskClarity = '$avg_task_clarity'";
if (mysql_query($print_task_clarity_1)) { echo $print_task_clarity_1; } else { echo "Error TaskClarity1: " . mysql_error();
First, mysql_query() returns a resource, and you then need to extract information from it. Your query doesn't mantion any table name (I'll call it MyTable).
Also, you can get all three averages with one query.
Here's how I would start:
$table = "MyTable";
$sql = "SELECT AVG(TaskClarity1) AS avgClarity1,
AVG(TaskClarity2) AS avgClarity2,
AVG(TaskClarity3) AS avgClarity1
FROM $table WHERE gruppid = '$group_id'";
$resource = mysql_query($sql); //execute the query
if (! $resource = mysql_query($sql) ){
echo "Error reading from table $table";
die;
}
if (! mysql_num_rows($resource ) ){
echo "No records found in $table";
}
else {
$row = mysql_fetch_assoc($resource); // fetch the first row
$avg_task_clarity_1 = $row['avgClarity1'];
$avg_task_clarity_2 = $row['avgClarity2'];
$avg_task_clarity_3 = $row['avgClarity3'];
$avg_task_clarity =
($avg_task_clarity_1+$avg_task_clarity_2+$avg_task_clarity_3)/3;
//...
// other stuff you want to do
}
Please comment if this is not helpful enough, and I will revise my answer.
here is the code that i am using, the code is simply getting the same value by two different methods, but the mysql_insert_id() echo is not giving the output.
The value given by -> echo mysql_insert_id(); is 0
Can anyone tell me the possible reason and how to resolve this problem?
include'config.php';
$query = "SELECT id
FROM users
ORDER BY id DESC
LIMIT 1";
if($query_run = mysql_query($query))
{
$query_result = mysql_fetch_row($query_run);
$id = $query_result[0];
echo $id;
echo $lastId = mysql_insert_id().'<br />';
//'Query Successful!';
} else {
echo mysql_error();
}
If you are looking for the next auto increment id that will be inserted. I guess you should try this.
$query_autoinc="SHOW TABLE STATUS LIKE 'mytable'";
$exec_autoinc=mysql_query($query_autoinc);
$row_autoinc=mysql_fetch_assoc($exec_autoinc);
$inserted_id = $row_autoinc['Auto_increment'];
Or if you want the last inserted id, you might try this.
$query= "SELECT max(id)
FROM users
ORDER BY id DESC
LIMIT 1";