I try querying very simple sql statement with mysqli
"select * from area where area_pre_id=6035;"
it returns nothing.
After querying this in phpmyadmin , it returns 78 rows ....
PHP code is as below;
$sql = "select * from area where area_pre_id=6035;";
if ($result = mysqli_query($conn, $sql, MYSQLI_USE_RESULT)) {
while($obj = $result->fetch_object()){
if($obj->area_local_name_th){
$my_province = $obj->area_local_name_th . "(" . $obj->area_eng_name . ")";
}else{
$my_province = $obj->area_eng_name;
}
$line[] = array("ProvinceID"=>$obj->area_id,"ProvinceName"=>$my_province);
}
}
Please tell me what's wrong with my code or sql statement.
Your mysqli command is right.I think there is no value in your database for that particular id.
Is the datatype of that id fild integer?
If it is integer then the query is right.But if it is varchar then you have to put a single quote.
select * from area where area_pre_id='6035';
You are trying to use both procedural and OOPs concept. THat will be the issue.
Try this
Change $result->fetch_object() to mysqli_fetch_object($result)
Related
I have a SQL query in my PHP file that makes use of some variables in it. I want to print the query itself on the localhost to check as to whether the entire query is been executed or not.
My query is like this:
$result = mysql_query("SELECT * FROM sample WHERE col01 LIKE '%$abc%',$db);
I am trying to print the query using echo $result but get Resource id #25 on localhost. I want to print Select * FROM ... as the output. Is there any way?
First of all: You are missing a double quote: $result = mysql_query("SELECT * FROM sample WHERE col01 LIKE '%$abc%'",$db).
That said, what stops you from
$sql="SELECT * FROM sample WHERE col01 LIKE '%$abc%'";
$result = mysql_query($sql,$db);
echo $sql;
If you were using PDO (and you should, the old mysql_ functions are deprecated and insecure) you could just use PDOStatement->queryString property to view the query at a later time.
Store as a variable $sql
Its normal, first you need to fetch that resource obj
And anyway you missing a double quote,
example.
$sql = "SELECT * FROM sample WHERE col01 LIKE '%$abc%'";
$result = mysql_query($sql);
while ($line = mysql_fetch_object($result)) {
echo $line->colname ."\n";
}
echo "\n" . ' query: ' . $sql
And from PHP 5.5.0 and beyond use mysqli
$sql = "SELECT * FROM sample WHERE col01 LIKE '%$abc%'";
if ($result = $mysqliobj->query($sql)) {
while($line= $result->fetch_object()){
echo = $line->colname ."\n";
}
}
echo "\n" . ' query: ' . $sql
or print_r($mysqliobj->info); # store las query
I am having problems trying to get these queries with a WHERE clause to work. I have two tables which look like this :
What I am trying to do is return the genre that each film has. At the moment no data is returning at all from what I can see. Here are the two queries:
$film_id = $row_movie_list['film_id'];
mysql_select_db($database_fot , $fot);
$query_get_genre = "SELECT * FROM film_genre WHERE `id_film` ='". $film_id. "'";
$get_genre = mysql_query($query_get_genre, $fot) or die(mysql_error());
$row_get_genre = mysql_fetch_assoc($get_genre);
$totalRows_get_genre = mysql_num_rows($get_genre);
$genre_id = $row_get_genre['id_genre'];
mysql_select_db($database_fot , $fot);
$query_genre = "SELECT * FROM genre WHERE `id_genre` ='". $genre_id. "'";
$genre= mysql_query($query_genre, $fot) or die(mysql_error());
$row__genre = mysql_fetch_assoc($genre);
$totalRows_genre = mysql_num_rows($genre);
PHP with content area. I fairly new to PHP so any help would be appreciated.
<?php do { echo $genre['genre']; } while($row_get_genre = mysql_fetch_assoc($get_genre)); ?>
Update: I am now able to get first genre but not second it just echos the first one twice and I have tried but still no luck:
do {do { echo $row_genre['genre']; } while($row_genre = mysql_fetch_assoc($genre));} while($row_get_genre = mysql_fetch_assoc($get_genre)); ?>
Avoiding the fact that you're using a deprecated way to establish connection and interact with MySQL, what you're doing is getting a single relation genre-film and then getting the row of the genre that matches. You should surround part of your code with a while that executes while it's still genres of the film with id. Something like:
$film_id = $row_movie_list['film_id'];
mysql_select_db($database_fot , $fot);
$query_get_genre = "SELECT * FROM film_genre WHERE `id_film` ='". $film_id. "'";
$get_genre = mysql_query($query_get_genre, $fot) or die(mysql_error());
while($row_get_genre = mysql_fetch_assoc($get_genre)){
$genre_id = $row_get_genre['id_genre'];
$query_genre = "SELECT * FROM genre WHERE `id_genre` ='". $genre_id. "'";
$genre= mysql_query($query_genre, $fot) or die(mysql_error());
$row__genre = mysql_fetch_assoc($genre);
// You should do whatever you want to do with $row__genre here. Otherwise it will be cleared.
}
I must insist this is a deprecated and insecure way of communication with a MySQL Database. I recommend you read about MySQLi or PDO extensions.
MySQLi: http://www.php.net/manual/en/book.mysqli.php
PDO: http://www.php.net/manual/en/book.pdo.php
This is the code I am using. It only returns queries whose value of $isbn does not start with 0.
The rows with isbn value like :09913456 are not returned.
The same query in PhpMyadmin works fine.
$isbn = $_GET["isbn"];
$query = 'SELECT * FROM crossword_data WHERE isbn LIKE '.$isbn;
$result = mysqli_query($con,$query);
while($row = mysqli_fetch_array($result))
{
echo $row['title'];
}
mysqli_close($con);
?>
Please be aware of sql-injection: the user data goes right into your database! escape it. Otherwise everybody could slowdown, read, write or empty your data.
You treat your input like a number, but you mean a string. A number doesn't start with 0.
Solution would be
'SELECT * FROM crossword_data WHERE isbn LIKE "'.$isbn . '"'
$query = 'SELECT * FROM crossword_data WHERE isbn LIKE '.'%".$isbn."%';
I have some code which selects data from a row. It is working and prints the result correctly, but I have an error I can not figure out.
$rs=array();
$select=array ("system_name", "location", "alarmtype", "severity", "start_time", "end_time", "duration", "reason","shift_oparation", "system_oparation");
for ($i=0;$i<=9;$i++){
$SQL = "SELECT (".$select[$i].") FROM (".$is.") WHERE duration=('".$ic."') AND location=('".$id."')";
$result = mysql_query($SQL);
$db_field = mysql_fetch_array($result);
$rs[$i]=$db_field[$select[$i]];
echo $rs[$i];
}
Echo prints correctly, but there is an error:
Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given.
I think some of your queries dont return anything. In order to get rid of the warning you should check if the $result was in fact a resource.
Try this:
$rs=array();
$select=array ("system_name", "location", "alarmtype", "severity", "start_time", "end_time", "duration", "reason","shift_oparation", "system_oparation");
for ($i=0;$i<=9;$i++){
$SQL = "SELECT (".$select[$i].") FROM (".$is.") WHERE duration=('".$ic."') AND location=('".$id."')";
$result = mysql_query($SQL);
if($result){
$db_field = mysql_fetch_array($result);
$rs[$i]=$db_field[$select[$i]];
echo $rs[$i];
}
}
However, your code will be a lot more efficient if you use only one query.
You are doing something really strange. It looks like $select is a list of columns in a table, and you are fetching them one column at a time!
You can select them all at once, with this query:
$sql = "SELECT system_name, location, alarmtype, severity,
start_time, end_time, duration, reason, shift_oparation, system_oparation
FROM " . $is . " WHERE duration = '" . $ic . "' AND location = '" . $id . "'";
Note I've removed some extraneous parenthesis you had. Also be very careful with building SQL queries dynamically like this. I'm assuming you know the values of $is, $ic, and $id to be safe from SQL injection attacks,
Run the result, and then iterate over the row array
$result = mysql_query($sql);
if ($result)
{
$row = mysql_fetch_array($result); //fetches the first result row as an array
// Can be accessed like $row['system_name']
foreach ($row as $key => $value)
echo $key, " = ", $value;
}
else
die(mysql_error() . $sql);
If the result is false, it will perform the die statement and print out the error returned from MySQL and also the query we built, so you can see if it looks correct.
Warning: mysql_fetch_array() expects parameter 1 to be resource,
boolean given in.
This means that mysql query produced errors .
Return Values
For SELECT, SHOW, DESCRIBE, EXPLAIN and other statements returning
resultset, mysql_query() returns a resource on success, or FALSE on
error.
Use mysql_error to display teh error and see why it happened.
e.g.
if(!$result){
echo mysql_error();
}
Protip: donlt use mysql libraries as noted in php manual
there is problem in select statement please check the select query or you can run individual select query and check the output
I have a problem with integers in MySQL. I am trying to update a cell which stores an integer. However, I have problem with the type of that cell. Actually it is type is set to int but when I retrive the data I always get 0 and I belive it is because of the problem about how I am trying to get it. Here is my code sample;
function updateNumb($username) {
$query = "SELECT `num` FROM `profiles` WHERE `nick`='" . $username . "'";
$result = mysql_query($query, $this->conn) or die(mysql_error());
$row = mysql_fetch_array($result, MYSQL_ASSOC);
$row['num'] = $row['num'] + 1;
$numb = (int)$row['num'] + 1;
//here I update the data.
$query = "UPDATE `profiles` SET `num`=" . $numb . " WHERE `nick`='".$username."'";
mysql_query($query, $this->conn) or die(mysql_error());
return $numb;
}
Can it be because of mysql_fetch_array stuff? Or how could I overcome this problem?
replace partisayisi with num
There is nothing wrong with the code you provided, maybe it's not doing what you really need, for example num is incremented twice, but there are no visible mistakes that would make it return 0, at least not in what we can see.
Make sure you provide valid username, try to echo your query before sending to mysql to see what it really looks like, maybe try this query yourself in mysql client or phpmyadmin to see what's going on.
Also if the only thing you need is to increment num for some user you can do it in one update, you don't need to use select to get that number:
UPDATE profiles set num=num+1 WHERE nick='somenick'