mysql_insert_id() not giving value - mysql

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";

Related

MYSQL Query fail to select possible result

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'];

Few warnings [expects parameter]

I've been trying to display the number of rows using this code but it keep says
1 Rows which is wrong
<?php
$link = mysql_connect("localhost", "gamin1_forum", "password123");
mysql_select_db("gamin1_forumdb", $link);
$result = mysql_query("SELECT COUNT(*) FROM smf_personal_messages", $link);
$num_rows = mysql_num_rows($result);
echo "$num_rows Rows";
?>
Rows are approximately 1443 but it kept saying 1
The Count(*) returns you one row which contains the number of rows as a value.
By using mysql_num_rows($result) you are actually counting the amount of rows of the Count(*) result which really is one.
Change it to:
$result = mysql_query("SELECT * FROM smf_personal_messages", $link);
$num_rows = mysql_num_rows($result);
Or just use the Count(*) value (which is probably better since it count in the DB and not retrieving the whole table for it) using mysql_fetch_array.
This is true, mysql_num_rows() is telling that there was one "row" returned. You need to check the value of the returned row to get your count. Try this
$link = mysql_connect("localhost", "gamin1_forum", "password123");
$result = mysql_query("SELECT COUNT(*) as cnt FROM smf_personal_messages", $link);
$row = mysql_fetch_array($result);
echo ($row['cnt']);
On a side note, you should not use mysql_* functions for security reasons. Try PDO/mysqli_* functions

MYSQL query omitting deleted

I have a column in my MySQL database for deleted values are usually 0 or 1. Usually when doing searches I omit the deleted by doing something like "and deleted = "0"" but I cant figure out how to get this query below to omit my deleted column. any ideas would be appreciated thanks!
$query = "SELECT Status, COUNT(Status) FROM Assets GROUP BY Status";
$result = mysql_query($query) or die(mysql_error());
// Print out result
while($row = mysql_fetch_array($result))
{
echo "". $row['COUNT(Status)'] ." systems ". $row['Status'];
echo ", ";
}
This should work:
$query = "SELECT Status, COUNT(Status) FROM Assets WHERE Assets.deleted = 0 GROUP BY Status";
Also, you should use mysqli or PDO rather than mysql (the php mysql library is obsolete).

Check whether value exist in database

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";
}

How to select multiple records (row) from a column in mysql?

I want to display four (4) items'name from these id:
Can I do like this?
SELECT item_name from items WHERE item_id IN ('001', '012', '103', '500')
or
SELECT item_name from items WHERE item_id = '001' or item_id = '012' or item_id = '103' or item_id = '500'
IN RESPONSE TO ALL ANSWERS
Well, most of the answers said it works, but it does not really work. Here is my code:
$query = "SELECT `item_name` from items WHERE item_id IN('s001','a012','t103','p500')";
$result = mysql_query($query, $conn) or die (mysql_error());
$fetch = mysql_fetch_assoc($result) or die (mysql_error());
$itemsCollected = $fetch['item_name'];
echo $itemsCollected;
The item_id is alphanumeric.
You can do either one, but the IN query is much more efficient for this purpose for any large queries. I did some simple testing long ago that revealed it's about 10 times faster to use the IN construct for this. If you're asking if the syntax is correct then yes, it looks fine, other than missing semi-colons to complete the statement.
EDIT: It looks like the actual question you were asking was "why do these queries only return one value". Well, looking at the sample code you posted, the problem is here:
$fetch = mysql_fetch_assoc($result) or die (mysql_error());
$itemsCollected = $fetch['item_name'];
echo $itemsCollected;
You need to loop through and iterate until there are no more results to be fetched, as Pax pointed out. See the PHP manual page for mysql_fetch_assoc:
$sql = "SELECT item_name from items WHERE item_id IN('s001','a012')";
$result = mysql_query($sql);
if (!$result) {
echo "Could not successfully run query ($sql) from DB: " . mysql_error();
exit;
}
if (mysql_num_rows($result) == 0) {
echo "No rows found, nothing to print so am exiting";
exit;
}
// While a row of data exists, put that row in $row as an associative array
// Note: If you're expecting just one row, no need to use a loop
// Note: If you put extract($row); inside the following loop, you'll
// then create $userid, $fullname, and $userstatus
while ($row = mysql_fetch_assoc($result)) {
echo $row["userid"];
echo $row["fullname"];
echo $row["userstatus"];
}
mysql_free_result($result);
Yes, both those should work fine. What's the actual problem you're seeing?
If, as you say, only one record is being returned, try:
select item_name from items order by item_id
and check the full output to ensure you have entries for 001, 012, 103 and 500.
If both those queries only return one row, I would suspect not.
If they all do exist, check the table definitions, it may be that the column is CHAR(4) and contains spaces for the others. You may have genuinely found a bug in MySQL but I doubt it.
After EDIT:
This is a perl/mysql problem, not an SQL one: mysql_fetch_array() returns only one row of the dataset at a time and advances a pointer to the next.
You need to do something like:
$query = "SELECT item_name from items WHERE item_id IN('s001','a012')";
$result = mysql_query($query, $conn) or die (mysql_error());
if (mysql_num_rows($result) == 0) {
echo "No rows found, nothing to print so am exiting";
exit;
}
while ($row = mysql_fetch_assoc($result)) {
echo $row["item_name"];
}
Your ID field must be set to auto increment, i guess. i had problems with that once and i changed the auto increment to int. in the IN field if you pass the parameters to match against the auto increment variable you get back only the first parameter, the remaining generates an error.
Use mysql_fetch_assoc to get the query and assign the values from mysql_fetch_assoc query into a an array. Simple as that
$i=0;
$fullArray = array();
$query = mysql_query("SELECT name FROM users WHERE id='111' OR id='112' OR id='113' ")or die(mysql_error());
while($row = mysql_fetch_assoc($query)){
foreach ($row as $value) {
$fullArray[$i] = $value;
}
$i++;
}
var_dump($fullArray);
echo $fullArray[0]."<br/>".$fullArray[1]."<br/>".$fullArray[2];`
You could also use the mysql_num_rows function to tell you how many rows your query retrieved and then use that result to increment a for loop. An example.
$num_rows=mysql_num_rows($query_results);
for ($i=0; $i <$num_rows ; $i++) {
$query_array[]=mysql_fetch_assoc($query_results);
}