Splitting a php array from mysql query - php

I have been working on this for hours on end trying to split the resultant mysql query into their usable variables and I just don't know what I am doing incorrectly. I no longer get the resource ID# but now all I get is "array" from the variables when I print_r. I am trying to get the 3 fields from the select statement in their own array and use the values. I have been pulling my hair out trying to get this to create usable data. Any help is greatly appreciated as I feel I could spend many, many more hours tinkering and just seem to be on the wrong track.
$Query = "SELECT guardian.Email, child.CFName, guardian.FName
FROM child
INNER JOIN uaccount ON child.ANum = uaccount.ANum
INNER JOIN guardian ON guardian.ANum = uaccount.ANum
WHERE uaccount.ANum = '$ANum';";
$result = mysql_query($query);
$test = mysql_fetch_array($result, MYSQL_BOTH);
print_r('this is the test '.$test."<br/>");
while($row = mysql_fetch_assoc($test, MYSQL_BOTH))
{
print_r('this is the row '.$row."<br/>");
foreach($row as $rows)
{
$info = mysql_fetch_assoc($rows, MYSQL_BOTH);
$Email = $info['0'];
$FName = $info['1'];
$CFName = $info['2'];
}
}
Thank you very much for your assistance thus far. With a combination of everyone's help I have been able to pull the values from the first level of the array with the following code :
$query = "SELECT guardian.Email, guardian.FName, child.CFName
FROM guardian
INNER JOIN uaccount ON child.ANum = uaccount.ANum
INNER JOIN guardian ON guardian.ANum = uaccount.ANum
WHERE uaccount.ANum = '$ANum';";
$result = mysql_query($query);
$info = mysql_fetch_array($result);
$Email = $info['0'];
$FName = $info['1'];
$CFName = $info['2'];
However, that is only returning me 1/3rd of the data as each account has 3 guardian email addresses related to it. Where I should be getting about 2000 rows returned, I am only getting 670. That is why I was nesting another layer inside the orginal posting and why I was attempting to pull a fetch_assoc from itself. If you cannot pull an array from itself, how do you "de-nest" so to speak? Any and all help is greatly appreciated.

$emQuery = "SELECT guardian.Email, child.CFName, guardian.FName
FROM child
INNER JOIN uaccount ON child.ANum = uaccount.ANum
INNER JOIN guardian ON guardian.ANum = uaccount.ANum
WHERE uaccount.ANum = '$ANum'";
$result = mysql_query($query);
$Email = array();
$FName = array();
$CFName = array();
$test = mysql_fetch_array($result, MYSQL_BOTH);
print_r('this is the test '.$test."<br/>");
while($row = mysql_fetch_assoc($test, MYSQL_BOTH))
{
echo 'this is the row ';
print_r($row);
echo '<br/>';
//$info = mysql_fetch_assoc($rows, MYSQL_BOTH);
$Email[] = $info['Email'];
$FName[] = $info['FName'];
$CFName[] = $info['CFName'];
}
To the best of my understanding this is what you are trying to do.
Some highlights:
print_r can only get a reference to a valid php array, no strings included.
mysql_fetch_assoc fetches a line out of a result set referenced by a variable, and then moves the reference to point to the next row.
You cannot call this function on a the result of it self, as it is not valid syntax.
In general, you'll be better off using PDO or mysqli_ functions at least, as it is by far more secure, by allowing you to use parameter binding instead of just using use input as part as your SQL.

You should just loop through the rows of your result once like so:
$result = mysql_query($query);
while ($row = mysql_fetch_assoc($result)) {
// print_r($row); Debug here if you want.
$Email = $row['Email'];
$FName = $row['FName'];
$CFName = $row['CFName'];
}
Note: You were providing a second parameter to mysql_fetch_assoc() which only takes one parameter (the result from a query). See the doc here.
mysql_fetch_array() takes another parameter that specifies what type of array to return.

Generalized PDO example out of a project of mine ($sql would be your $Query string):
$qry = $pdo->query($sql);
$all=array();
$idx=0;
while($fields=$qry->fetch( 'PDO::FETCH_ASSOC' )){
$all[key($fields)][$idx]=$fields[key($fields)];
while(next($fields)!==false){
$all[key($fields)][$idx]=$fields[key($fields)];
}
$idx++;
}
The $all variable will hold an array which in turn holds an array for each of the columns you've selected.
Like that you can do neat things like array_combine($all['Email'],$all['Fname']) and you'd get an array where the key is the Email column and the value would consist of the Fname column.

Related

Create php array from one of all column that retrieve from mysql

I am a newbie on PHP,MySQL and HTML.
I have one question about php array.
I create MySQL query that joins 2 tables in php. From this join, I will get information about one product, but have multiple JIG.
$sql = "SELECT product.product_number,product.product_name,product.product_jitqty,product.product_desc,jit.jit_number,jit.jit_name,jit.jit_drawer,jit.jit_port,jit.jit_specpath
FROM product
JOIN production_jit
ON production_jit.product_number = product.product_number
JOIN jit
ON jit.jit_number = production_jit.jit_number
WHERE product.product_number = '$productnumber'";
$result = mysqli_query($conn, $sql);
if (mysqli_num_rows($result) > 0) {
while($row = mysqli_fetch_assoc($result))
{ }
from above query, I will get result like picture below.
So I want to create an array for column jit.jit_number. for easy to transfer both or only one data to another page.
I have tried below code.
$sql = "SELECT product.product_number,product.product_name,product.product_jitqty,product.product_desc,jit.jit_number,jit.jit_name,jit.jit_drawer,jit.jit_port,jit.jit_specpath
FROM product
JOIN production_jit
ON production_jit.product_number = product.product_number
JOIN jit
ON jit.jit_number = production_jit.jit_number
WHERE product.product_number = '$productnumber'";
$result = mysqli_query($conn, $sql);
if (mysqli_num_rows($result) > 0) {
// output data of each row
?>
<?php
while($row = mysqli_fetch_assoc($result))
{
$jig = array ($row['jit_number']);
echo "This is JIG " . $jig[0] . ", " . $jig[1] . ".";
}
But the result is like picture below.
Can anyone help me? So if I want to transfer both data to another page, I just can use $jig[0] and $jig[1]. Or anyone can advise me the better way to transfer multiple data form mysql to another page.
I'm not really sure what you're trying to achieve, but $jig is not going to have two indices because $row['jit_number'] ist just a number itself. If you're trying to store all jit_numbers in a separate array, you could try something like this:
$numbers = array();
while($row = mysqli_fetch_assoc($result))
{
$numbers[] = $row['jit_number'];
}
You shall also try to prevent MySQL injection, WHERE product.product_number = '$productnumber'"; is rather horrible from a security point of view. Using a prepared statement or at least some sanitization / escaping is recommendable.
Try this->
$jig = array();
array_push($jig, $row['jit_number']);

How can I get this php to return the entire column of an sql db

I am trying to query a db for an entire column of data, but can't seem to get back more than the first row.
What I have so far is:
$medicationItem = array();
$medicationItemSql = "SELECT medication FROM medication";
$medicationItemObj = mysqli_query($connection, $medicationItemSql);
if($row = mysqli_fetch_array($medicationItemObj, MYSQLI_NUM)){
echo count($row);
}
It's not my intention to just get the number of rows, I just have that there to see how many it was returning and it kept spitting out 1.
When I run the sql at cmd line I get back the full result. 6 items from 6 individual rows. Is mysqli_fetch_array() not designed to do this?
Well, I had a hard time understanding your question but i guess you are looking for this.
$medicationItem = array();
$medicationItemSql = "SELECT medication FROM medication";
$medicationItemObj = mysqli_query($connection, $medicationItemSql);
if($row = mysqli_num_rows($medicationItemObj))
{
echo $row;
}
Or
$medicationItem = array();
$medicationItemSql = "SELECT medication FROM medication";
$medicationItemObj = mysqli_query($connection, $medicationItemSql);
$i = 0;
while ($row = mysqli_fetch_array($medicationItemObj))
{
$medicationItem[] = $row[0];
$i++;
}
echo "Number of Rows: " . $i;
If you just want the number of rows i would suggest using the first method.
http://php.net/manual/en/mysqli-result.num-rows.php
You can wrote your code like below
$medicationItem = array();
$medicationItemSql = "SELECT medication FROM medication";
$medicationItemObj = mysqli_query($connection, $medicationItemSql);
while ($row = mysqli_fetch_assoc($medicationItemObj))
{
echo $row['medication'];
}
I think this you want
You could give this a try:
$results = mysqli_fetch_all($medicationItemObj, MYSQLI_NUM);
First, I would use the object oriented version of this and always use prepared statements!
//prepare SELECT statement
$medicationItemSQL=$connection->prepare("SELECT medication FROM medication");
// execute statement
$medicationItemSQL->execute();
//bind results to a variable
$medicationItemSQL->bind_result($medication);
//fetch data
$medicationItemSQL->fetch();
//close statement
$medicationItemSQL->close();
You can use mysqli_fetch_assoc() as below.
while ($row = mysqli_fetch_assoc($medicationItemObj)) {
echo $row['medication'];
}

getting results of SQL query

I've seen there there are numerous ways to put the results of a SQL query into usable format (that is, variables).
If I have a targeted SQL query that I know will be returning a set of expected values, lets say querying a customer number to pull data, city, state, first and last name, etc..
Code example follows:
$example = '50';
$result = mysql_query("SELECT * FROM customers WHERE customer_ID = '$example'",$db);
while ($row = mysql_fetch_assoc($result)) {
foreach ($row as $col => $val) {
if ($col == 'firstname') {
$customerfirstname = $val;
}
}
}
or another way:
$result = mysql_query("SELECT * FROM customers WHERE customer_ID = '$example'",$db);
$myResultArray = array();
while ($row = mysql_fetch_assoc($result))
$myResultArray = $row;
foreach ($myResultArray as $val) {
$customerfirstname = $val['firstname'];
}
That's just two that I could think of.
Is one of the above methods preferable over the other? If so, why?
Is there an alternate method that is even more efficient than either of these?
Neither are preferred.
The foreach's are superfluous.
Since you know the fieldnames you need, you can just do:
while ($row = mysql_fetch_assoc($result)) {
$customerfirstname = $row['firstname'];
}
If you do need to apply a conditional for some reason, you can test for the field's existence in the array:
while ($row = mysql_fetch_assoc($result)) {
if (isset($row['firstname'])) {
$customerfirstname = $row['firstname'];
}
}
Finally, since you appear to be selecting by primary key, the while loop is also unnecessary:
if ($row = mysql_fetch_assoc($result)) {
$customerfirstname = $row['firstname'];
}
I have used the first example that you posed in every website I have done that requires a database and it hasn't failed me yet. As far as if one is better than the other I'd say no. It's just a matter of taste.
There are many ways. Here's a quick one, but I would prefer to set it up using a DTO and accessing it that way... this will work though for your question.
$query = "SELECT first_name, last_name, city FROM customers WHERE customer_id = '$example'";
$result = mysql_query($query);
// If you are expecting only one row of data, then use this:
list($first_name, $last_name, $city) = mysql_fetch_row($result);
//several rows:
while(list($first_name, $last_name, $city) = mysql_fetch_row($result)){
echo $first_name;
}
I seem to be missing something...
Why not this?
$result = mysql_query("SELECT * FROM customers WHERE customer_ID = '$example'",$db);
while ($row = mysql_fetch_assoc($result)) {
$customerfirstname = $row['firstname'];
}
In the first example?

Get rows from mysql table to php arrays

How can i get every row of a mysql table and put it in a php array? Do i need a multidimensional array for this? The purpose of all this is to display some points on a google map later on.
You need to get all the data that you want from the table. Something like this would work:
$SQLCommand = "SELECT someFieldName FROM yourTableName";
This line goes into your table and gets the data in 'someFieldName' from your table. You can add more field names where 'someFieldName' if you want to get more than one column.
$result = mysql_query($SQLCommand); // This line executes the MySQL query that you typed above
$yourArray = array(); // make a new array to hold all your data
$index = 0;
while($row = mysql_fetch_assoc($result)){ // loop to store the data in an associative array.
$yourArray[$index] = $row;
$index++;
}
The above loop goes through each row and stores it as an element in the new array you had made. Then you can do whatever you want with that info, like print it out to the screen:
echo $row[theRowYouWant][someFieldName];
So if $theRowYouWant is equal to 4, it would be the data(in this case, 'someFieldName') from the 5th row(remember, rows start at 0!).
$sql = "SELECT field1, field2, field3, .... FROM sometable";
$result = mysql_query($sql) or die(mysql_error());
$array = array();
while($row = mysql_fetch_assoc($result)) {
$array[] = $row;
}
echo $array[1]['field2']; // display field2 value from 2nd row of result set.
The other answers do work - however OP asked for all rows and if ALL fields are wanted as well it would much nicer to leave it generic instead of having to update the php when the database changes
$query="SELECT * FROM table_name";
Also to this point returning the data can be left generic too - I really like the JSON format as it will dynamically update, and can be easily extracted from any source.
while ($row = mysql_fetch_array($result, MYSQL_ASSOC))
{
echo json_encode($row);
}
You can do it without a loop. Just use the fetch_all command
$sql = 'SELECT someFieldName FROM yourTableName';
$result = $db->query($sql);
$allRows = $result->fetch_all();
HERE IS YOUR CODE, USE IT. IT IS TESTED.
$select=" YOUR SQL QUERY GOOES HERE";
$queryResult= mysql_query($select);
//DECLARE YOUR ARRAY WHERE YOU WILL KEEP YOUR RECORD SETS
$data_array=array();
//STORE ALL THE RECORD SETS IN THAT ARRAY
while ($row = mysql_fetch_array($queryResult, MYSQL_ASSOC))
{
array_push($data_array,$row);
}
mysql_free_result($queryResult);
//TEST TO SEE THE RESULT OF THE ARRAY
echo '<pre>';
print_r($data_array);
echo '</pre>';
THANKS

is_array for MySQL Select Statement with PHP

I have a basic PHP function that I am working with. Sometimes, it is passed an array of variables, and other times it is just passed one variable. So, currently, I have something like this:
<?
function do_this($user_id_array) {
$user_ids = array();
foreach ($user_id_array as $single_user_id) {
$sql = 'SELECT username FROM users
WHERE id = $single_unit';
while($row = mysql_fetch_assoc($result)) {
array_push($user_ids, $row['id'];
}
}
return $user_ids;
}
?>
My issue is this: If I call the function and send it only one variable (and not an array), it (obviously) gives me the following error: Invalid argument supplied for foreach()
My question is: How can I change this function in the most efficient way with the least amount of code? Do I have to use an if is_array() statement and just create 2 SELECT statements, one for each case (array and non-array)?
Many thanks!
I see several options:
Pass an array even if it's one element long
Test for is_array() and act accordingly
Add another argument which states whether to check for an int or an array.
I'd go with options 1 or 2, as option 3 is error prone.
Also, there might be a better solution to your problem, you shouldn't have a single query for every user, you should instead use the IN keyword in MySQL, something like this:
$users = (is_array($user_id_array)) ? implode(',',$user_id_array) : $user_id_array;
$query = "SELECT `username` FROM `users` WHERE `id` IN({$users})";
wow. that's a lot of queries. What about to delete foreach and do something like
if (is_array($user_id_array)){
$sql = 'SELECT username,id FROM users
WHERE id IN ('.implode(",", $user_id_array).')';
$result = mysql_query($sql);
while($row = mysql_fetch_assoc($result)){
$users[$row['id']] = $row;
}
}
You can write:
$user_id_array = (array)$user_id_array;
if (!is_array($user_id_array))
$user_id_array = array($user_id_array);
function do_this($user_id_array) {
$ids = array_map('intval', (array)$user_id_array);
$sql = 'SELECT username FROM users
WHERE id IN(' . implode(',', $ids) . ')';
$result = mysql_query($sql);
$usernames = array();
while ($row = mysql_fetch_assoc($result)) {
$usernames[] = row['id'];
}
return $usernames;
}
First line makes sure that you have an array ((array)$user_id_array) and that all values are valid integers. Then a single SQL query is executed for all user ids.

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