Im trying to pass an array that I already found by a query into another query. For example:
$first_query = "SELECT id FROM from Table WHERE user = '{$_SESSION['id'}'";
$result = mysql_query($first_query,$connection);
$ids = mysql_fetch_array($result);
This is where it gets tricky for me. I want to pass $ids into the next query.
$id_implode = implode(", ", $ids)
$second_query = "SELECT * FROM Table2 WHERE id = '{$id_implode}';
The second query doesnt seem to be working. Any help is greatly appreciated!
your second query's syntax is wrong. Once evaluated it should read
select * from Table2 where id in (1,2,3)
ditch the curly braces and change the = to in. Don't use OR - that's a dumb way of ignoring good sql functionality
EDIT: Teneff's comment makes a very good point - why are you approaching the problem in this way? If there is a relationship between the tables they can be joined and all the data you want can be retrieved in a single query. If for some reason you can't / won't join the tables you could at least try a sub-query
select * from table2 where id in (select id from table where user = $_SESSION['id']);
To use a where statement with multiple entries to match on, use in ().
$id_implode = "'".implode("', '", $ids)."'"
$second_query = "SELECT * FROM Table2 WHERE id in ({$id_implode});
I think you should use IN
$id_implode = implode(", ", $ids)
$second_query = "SELECT * FROM Table2 WHERE id IN '({$id_implode})';
This assumes that $ids is made of int of course, otherwise you have to enclose eache entry in quotes. that means
IN (6,7,8,9)//this doesn't need quotes
IN ('lemon', 'orange')//needs quotes
try to use the IN syntax:
$id_implode = implode("', '", $ids);
$second_query = "SELECT * FROM Table2 WHERE id in ('{$id_implode}');
Related
My table look like this:
Table screenshot
Here I'm getting the result by query:
$subject_ids = implode(',', $_POST['subject_ids'])
SELECT * FROM table WHERE focusarea LIKE '%$subject_ids%' ;
The result is perfect, but there is nothing to display when I select more than one subject ids, like if selecting only one then it shows,
but when to select 1, 2, and 4, but there is nothing with this LIKE query...
How can I fix this?
Use implode like,
PHP
$subject_id_aray = explode(",",$_POST['subject_ids']);
$in_array_string = array();
foreach($subject_id_aray as $values){
$in_array_string[] = "'".$values."'";
}
MySql
$sql = "SELECT * FROM table WHERE focusarea in (".implode(",",$in_array_string).") ;";
LIKE clause will not work in your case because using LIKE '%1,2,3%' in query will not get anything, as you as using Ids you should use IN instead of LIKE. LIKE will be used separately for each id if it is string.
As you are getting $_POST['subject_ids'] as an array, query will be like
$subject_str = implode(',', $_POST['subject_ids']);
$sql = "SELECT * FROM table WHERE focusarea IN($subject_str)";
If your column focusarea is not integer then
$subject_str = "'".implode("','", $_POST['subject_ids'])."'";
$sql = "SELECT * FROM table WHERE focusarea IN($subject_str)";
Maybe you have bug in POST.
Try to echo, $subject_ids befor inject to SQL.
You focus are is simple string of numbers, connected by ,, but what you are sending by POST maybe is not correct.
Other problem, this don't look like you full code.
Provide you file, if this don't resolve problem.
How do I make it pick all results that are not equal to the $var , here's my code.
$opti=mysql_query("SELECT * FROM table1 WHERE imageid=$image_id");
while ($vari = mysql_fetch_array($opti)) {
$var = $vari['tagid'];
$options=mysql_query("SELECT * FROM table WHERE id!=$var");
while ($taghe1 = mysql_fetch_array($options)) {
$tagname = $taghe1['name'];
echo "".$tagname.", ";
} }
Try:
$options=mysql_query("SELECT * FROM table WHERE id<>{$var}");
You can probably see from the answer you accepted that adding the quotes solved your problem. Another way to do this is to just use one query. I will show an example using mysqli instead of the deprecated mysql, but the same query should work in mysql if you must use it. I added a couple of other suggestions that aren't really addressing your question, but make me feel better about my answer.
// Please be sure to escape $image_id before using it like this
$unused_tags = mysqli_query($db, "SELECT `name` FROM `table` AS t
LEFT JOIN (SELECT tagid FROM table1 WHERE imageid=$image_id) AS t1
ON t.id = t1.tagid WHERE t1.tagid IS NULL;");
while ($tag = mysqli_fetch_array($unused_tags)) {
$tags[] = htmlspecialchars($tag['name']); // escape your output
}
echo implode(", ", $tags); // doing it this way eliminates the trailing comma
You could use this:
$options=mysql_query("SELECT * FROM table WHERE id not in ('$var')");
You could have multiple values here, e.g.
$options=mysql_query("SELECT * FROM table WHERE id not in ('$var1', '$var2', '$var3')");
In my table, I have a column that it has a set of users id. its name is users_id.
each id separated with a comma(,).
ex : users_id = '1,2,3,4,5';
if I passed $id=1 to my function, how can I using where statement ?
function($id){
$sql = "select * from content_noti
where ??????"
}
You can add a leading and a trailing , in the field value in DB.
e.g.
Change
1,2,3,4,5
to
,1,2,3,4,5,
So that every Id has leading and trailing comma ,
Now, update the function body:
function($id){
$sql = "select * from content_noti
where users_id LIKE '%,$id,%'"
}
In this case you don't have any possibility of mistake as if you search user id 1 then you will get only 1 and not 11 or 111 or 1343.
Working Demo
if you have multiple ids than it would be better to use IN with query
$sql = "select * from content_noti where id IN (?)"
if you are looking for reverse than use find_in_set
$sql = "select * from content_noti where FIND_IN_SET(?, id)";
You can use regular expressions in mysql:
function($id){
$sql = "SELECT * FROM content_noti WHERE users_id REGEXP '(^|,)" . $i . "($|,)'";
}
SELECT * from content_noti where users_id = "1" OR users_id = LIKE '%1,%' OR users_id = LIKE '%,1%'
Storing multiple chunks of data as a single comma seperated colum is really poor design for databases and if at all possible you should look into normalizing this by making a coupling table and joining on that.
The performance of the above query will be terrible and it'll be hard to maintain.
I am trying to do a query in PHP PDO where it will grab a simple result. So like in my query I need it to find the row where the column group is 'Admin' and show what ever is in the group column. I know that we already know what it should be [Should be admin] but just need to get the query to work. Its only grabbing 1 row from my table, so will I need forsearch?
If I change WHERE group = 'Admin' to WHERE id = '1' it works fine. But I need it so it can be where group = 'admin'
$sql2 = "SELECT * FROM groups WHERE group = 'Admin'";
$stm2 = $dbh->prepare($sql2);
$stm2->execute();
$users2 = $stm2->fetchAll();
foreach ($users2 as $row2) {
print ' '. $row2["group"] .' ';
}
Thanks
group is a reserved word in MySQL, that's why it's not working. In general it's a bad idea to use reserved words for your column and table names.
Try using backticks around group in your query to get around this, so:
$sql2 = "SELECT * FROM groups WHERE `group` = 'Admin'";
Also you should really use placeholders for values, because you're already using prepared statement it's a small change.
Edit: just to clarify my last remark about the placeholders. I mean something like this:
$sql2 = "SELECT * FROM groups WHERE `group` = ?";
$stm2->execute(array('Admin'));
try to use wildcard in your WHERE Clause:
$sql2 = "SELECT * FROM groups WHERE group LIKE '%Admin%'";
Since the value in your table is not really Admin but Administrator then using LIKE and wildcard would search the records which contains admin.
For the user I am testing with, their org_id column value is "student_life"
I am trying to have this function display whatever rows have the student_life column = 1. (so yes there is a column student_life which is a boolean, and then I also have a separate column named org_id and in this case has the value student_life)
I am pretty sure there is a syntax error but I cannot figure it out.
function org_id_users_table()
{
$org_id = mysql_real_escape_string($_POST["org_id"]);
$sql = $this->query("SELECT * FROM ".DBTBLE." WHERE '$org_id' = '1'");
$result = $sql['sql'];
$num_rows = $sql['num_rows'];
$this->create_table($result, $num_rows);
}
(when I replace $org_id in the "$sql=..." line with student_life the code works.
You're quoting the column name, which makes MySQL think it's a string.
$sql = $this->query("SELECT * FROM ".DBTBLE." WHERE $org_id = '1'");
Edit:
Based on your comments, I think what you actually want is this:
$sql = $this->query("SELECT * FROM ".DBTBLE." WHERE org_id = '$org_id'");
Change quotes.
$sql = $this->query("SELECT * FROM ".DBTBLE." WHERE `$org_id` = '1'");
P.S. Why shouldn't I use mysql_* functions in PHP?
Where is this coming from? $_POST["org_id"]
Do you have a form on the page posting that? Or are you just trying to get that from the database? If so, wouldn't you need another query to obtain that first?
$row_MyFirstQuery['org_id']
Otherwise if it is $_POST["org_id"], wouldn't it be single quotes not double? $_POST['org_id']