MYSQL Statement with Array Variable - php

So I have the following mysql statement saving to an array:
$sql = "SELECT did FROM did WHERE id = '$did_id'";
$result = mysqli_query($link, $sql, MYSQLI_STORE_RESULT);
while($row = $result->fetch_row()){
$did[] = $row;
}
That part works great. But now I need to take the values in the $did array and perform another lookup using the values in it. The way it works is we have users assigned to certain did's. So I find the did's that the user is assigned to (the $did array) and only show them results from another table based on those did values. I have no idea how this part works, but this is what my next statement needs to do:
SELECT * FROM log WHERE did_id = "the values in $did array"
Hope someone can help. I really appreciate it. I haven't really been able to find anything on it.

You can use php's join with mysql's IN to make comma separated strings you can also use implode , join() is an alias of implode();
SELECT * FROM log WHERE did_id IN( ".join(',',$did).")
One thing to mention here that your $did should contains ids in manner like
array("1","2","3"....)
So in your loop fetch first index that holds did column value
$did[] = $row[0];
Note: i assume did , did_id type is int or bigint

Related

Notice: Array to string conversion. without array in code

the problem=
in my code i get the id of a person out of the database, then i fetch it and then i put it on another table in the database.
so first this i get the id out of the database:
$patientid = mysqli_query($connection, "SELECT id FROM patient WHERE `naam` = '$naam' AND `adres` = '$adres'");
then i fetch it:
$patientid1 = mysqli_fetch_assoc($patientid);
then i put it in another table:
$toevoegen = mysqli_query($connection, "INSERT INTO factuur (`soort`, `hoelang`, `titel`, `omschrijving`, `datum_aangemaaktfactuur`, `patient_id`, `artsid`)
VALUES ('$soort', '$hoelang', '$titel', '$omschrijving', '$datum', '$patientid1', '$artsid1')");
then when i run the code i get this notice: Notice: Array to string conversion.
how do i solve this? i have looked at other questions but they have the same error but different code. i dont even have an array in my code!
thanks
ps: if i replace $patientid1 with an number it works perfectly.
mysqli_fetch_assoc returns an associative array with selected fields as keys and column values as values.
You should modify code as follows
$patientRow = mysqli_fetch_assoc($patientid);
$patientId = $patientRow['id'];
Remember that for non unique results, you should iterate over whole result set like
while ($patientRow = mysqli_fetch_assoc($patientid)) {
$patientId = $patientRow['id'];
// other actions here
}
Last but not least, pay attention to $naam and $naam: use prepared statement instead will be definitely safer

Comparing a PHP variable to values in MySQL column

Just a little assistance, This is a pretty simple problem but it doesn't seem to work right. I am just comparing the value in a variable with all the values in a sql column. Same as if I were to compare a username input to the list of usernames in a sql column. This however is just to compare that the item id being stored in the column for that row is not an item id that is already in use.
I tested the value that I am getting back from the sql query and it is equal to the item id I typed in the input. What you will see below is the actual test to see if the id I am getting back is the one that I am looking for as well as the id of the row I can find that value in. The results I get is
2, 000002 (which is correct) that is what I am looking for.
$itemId = $_POST['itemId'];
if($sqlItemId = $dbCon->query("SELECT * FROM CVCinStoreCoins WHERE itemId = '$itemId'")){
while($data = $sqlItemId->fetch_assoc()){
printf("<p>%s, %s</p>", $data['id'], $data['itemId']);
die();
}
Then I took this out and tried to compare the value in the variable which is the same itemId already stored (000002). that is where I am going wrong.
I modified the code to look like this for further testing. Seems straight forward yet i am getting a FALSE response providing the latter echo statement "Item Id is not in use" But it is in the DB. I tried it a few different ways based on what I read in stackoverflow but none are giving me the right answer.
$sqlItemId = $dbCon->query("SELECT * FROM CVCinStoreCoins WHERE itemId = '$itemId'");
if($itemId == $sqlItemId){
echo "This item id is already in use. \n";
die();
} else {
echo "Item Id is not in use:";
die();
}
At one point I even tried a while statement to fetch the associated values prior to testing it but that didn't turn up a positive result either. Any suggestions?
Inside $sqlItemId you have the full table row (if any), not only its ID; change the SQL into a count and check the number of rows returned (if greater than 0 you have a duplicate):
$rowsCount = $dbCon->query("
SELECT COUNT(*)
FROM CVCinStoreCoins
WHERE itemId = '$itemId'
");
I don't know what $dbCon is (Doctrine DBAL? mysqli?) so I can't tell you how to use query's result.
Wy don't you just count it,
$result = $dbCon->query("SELECT COUNT(itemId) FROM CVCinStoreCoins WHERE itemId = $itemId");
if $result > 0

passing parameter to select statement from url

I have a database with a table which has two columns, lets say aa_id and bb_id - each of the the columns is a foreign key relating to another table and both columns are making a composite key for this particular table. there are several rows containing either the same aa_id and different bb_id or the same bb_id and different aa_id.
using pdo I want to extract the rows of the same - let's say - aa_id and I want to do this passing the parameter value in url. so the result of the select statement should be several rows and they should be saved as - for example - an array.
I have tried to do this with following code:
$sql = sprintf("select aa_id, bb_id from a_table where aa_id=:aa_id");
$res = $db->query($sql);
$rows = $res->fetch(PDO::FETCH_ASSOC);
foreach($rows as $key=>$value)
{
echo $key . " - " . $value . "</br>";
}
And it give no result.
I does work if I state the value of aa_id in the query like this
$sql = sprintf("select aa_id, bb_id from a_table where aa_id=191919");
, but it extracts no data if I put the value in url.
I am not really sure what to search for in the web because I don't know what's the notation called (if it is). If somebody could tell me what may be wrong with the code or give me directions to what I should look for in the web among tutorials or documentation I will be grateful. Perhaps somebody could recommend a good source of knowledge about mysql, php and pdo... Thanks in advance.
Well yeah, :indicator doesn't just automatically load in $_GET['indicator'], you need to manually bind it.
Assuming the URL ends with, ?aa_id=191919, your code might look something like this:
$sql = "select aa_id, bb_id from a_table where aa_id=:aa_id";
$res = $db->prepare($sql);
$res->bindValue(':aa_id', $_GET['aa_id'], PDO::PARAM_INT);
$res->execute();
$rows = $res->fetch(PDO::FETCH_ASSOC);
while($row=$res->fetch(PDO::FETCH_ASSOC))
{
print_r($row);
}

How to query all fields in a row

I know this is very simple, but I haven't used PHP/MySQL in a while and I have been reading other threads/php website and can't seem to get it.
How can I query a single row from a MySQL Table and print out all of the fields that have data in them? I need to exclude the NULL fields, and only add those that have data to an html list.
To clarify, I would like to display the field data without specifying the field names, just for the reason that I have a lot of fields and will not know which ones will be NULL or not.
What you've outlined requires 4 basic steps:
Connect to the database.
Query for a specific row.
Remove the null values from the result.
Create the html.
Step 1 is quite environment specific, so that we can safely skip here.
Step 2 - SQL
SELECT * from <tablename> WHERE <condition isolating single row>
Step 3 - PHP (assuming that $query represents the executed db query)
//convert the result to an array
$result_array = mysql_fetch_array($query);
//remove null values from the result array
$result_array = array_filter($result_array, 'strlen');
Step 4 - PHP
foreach ($result_array as $key => $value)
{
echo $value \n;
}
Just SELECT * FROM table_name WHERE.... will do the trick.
To grab data from specific fields, it would be SELECT field_1,field_2,field_3....
you have to make a string which represent mysql query. Then there is function in php named mysql_query(). Call this function with above string as parameter. It will return you all results. Here are some examples
You need to do it like this...
First connect to your sql... Reference
Now make a query and assign it to a variable...
$query = mysqli_query($connect, "SELECT column_name1, column_name2 FROM tablename");
If you want to retrieve a single row use LIMIT 1
$query = mysqli_query($connect, "SELECT column_name1, column_name2 FROM tablename LIMIT 1");
If you want to fetch all the columns just use * instead of column names and if you want to leave some rows where specific column data is blank you can do it like this
$query = mysqli_query($connect, "SELECT * FROM tablename WHERE column_name4 !=''");
Now fetch the array out of it and loop through the array like this..
while($show_rows = mysqli_fetch_array($query)) {
echo $show_rows['column_name1'];
echo $show_rows['column_name2'];
}
If you don't want to include the column names in the while loop, you could do this:
while($show_rows = mysqli_fetch_array($query)) {
foreach( $show_rows as $key => $val )
{
echo $show_rows[$key];
}
}

How do i select multiple rows which match the value of an item in an array?

I have written a php script that returns an arbitrary number of specific ids (which are in the format of numbers) in an array. I would like to make a new query that selects each row from a table that belongs to each id. I know i can do 1 query to get one row with the matching id. But i would like to do this all in one query. Here is my code:
$id = array(1,4,7,3,11,120); // The array containing the ids
$query = mysql_query("SELECT *
FROM posts
WHERE id = '$id[0]'");
// I would like to Do this for each id in the array and return it as one group of rows.`
I think you want the IN clause:
$idList = implode(",", $id);
SELECT *
FROM posts
WHERE id IN ( $idList );
The implode() function will turn your array of numbers into a comma-separated string of those same values. When you use it as part of an IN clause, it tells the database to use those values as a lookup table to match id against.
Standard Disclaimer/Warning:
As with any SQL query, you really shouldn't be directly concatenating variables into the query string. You're just opening yourself up to SQL injection. Use prepared/parameterized statements instead.
Use PHP's implode function to convert the array into a comma separated value string.
Then, you can use the SQL IN clause to run a single SQL statement containing the values associated with the ids you captured from PHP:
$id = array(1,4,7,3,11,120);
$csv = implode(',', $id);
$query = sprintf("SELECT *
FROM posts
WHERE id IN (%s)",
mysql_real_escape_string($csv));
$result = mysql_query($query)
I omitted the single quotes because they aren't necessary when dealing with numeric values in SQL. If the id values were strings, each would have to be encapsulated inside of single quotes.
What you want is SQL's IN clause.
SELECT * FROM posts WHERE id IN (1, 4, 7, 11, 120)
In PHP, you'll probably want something like:
$query = mysql_query(sprintf("SELECT * FROM posts WHERE id IN (%s)", implode(',', $id)));
Obviously, that's assuming you know you have integer values for $id, and that the values for $id didn't come from the user (that is, they should be sanitized). To be safe, you really ought to do something like:
$ids = implode(',', array_map('mysql_real_escape_string', $id));
$query = mysql_query("SELECT * FROM posts WHERE id IN ($ids)");
And if $id is dynamically generated, don't forget to put something in that IN clause, because SELECT * FROM foo WHERE bar IN () will give you an error. I generally make sure to set my IN-clause variables to 0, since IN (0) is good, and primary keys are pretty much never 0.

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