I have a query in mysql which will return a table grouped by some fields. How can I echo these results?
Here is piece of my code.
$stmt = $con->prepare("
SELECT classSubject
, COUNT(*) cnt
from Class
where grade=?
and department=?
and classGroup IN(?,3)
GROUP
BY classSubject;
");
$stmt->bind_param("isi", $grade, $department, $classGroup);
//executing the query
$stmt->execute();
What should I do after this so that I can echo all the result?
I recommend reviewing a good site which can serve as an example of what syntax to use, such as this one.
$sql = "SELECT classSubject, COUNT(*) cnt FROM Class ";
$sql .= "WHERE grade = ? AND department = ? AND (classGroup = ? OR classGroup = 3) ";
$sql .= "GROUP BY classSubject";
$stmt = $mysqli->prepare($sql);
$stmt->bind_param("isi", $grade, $department, $classGroup);
$stmt->execute();
$result = $stmt->get_result();
if ($result->num_rows > 0) {
while ($row = $result->fetch_assoc()) {
echo "(classSubject, count) = (" . $row['classSubject'] . ", " . $row['cnt'] . ")";
}
}
Note that I add an alias to the COUNT(*) term in the select so that we can easily refer to it by name in PHP. You could also just refer to columns using their numeric indices, but going by name is an easier and safer way to do it.
After reading Tim Biegeleisen's answer and doing a bit of surfing, I came up with this solution. After executing the query:
$response = array();
$response['success']=true;
$stmt->bind_result($classSubject,$cnt);
while ($stmt->fetch())
{
$response[$classSubject]=$cnt;
}
echo json_encode($response);
Hope this helps someone.
Related
I have a database have 20 tables I want to search all of these I got stuck that how can I search this and solve this query help is very appreciated. Like:
table1
table2
table3
table4
and so one. This is my script.
<?php
if(isset($_GET["search"]))
{
$condition = '';
//$query = explode(" ", $_GET["search"]);
$query = explode(" ", $_GET["search"]);
foreach($query as $text)
{
$condition .= "`title` LIKE +'%".mysqli_real_escape_string($connect, $text)."%' OR ";
}
$condition = substr($condition, 0, -4);
$sql_query = "SELECT * FROM countries2 WHERE " . $condition;
$result = mysqli_query($connect, $sql_query);
if(mysqli_num_rows($result) > 0)
{
while($row = mysqli_fetch_array($result))
{
echo '<tr><td>'.$row["title"].'</td></tr>';
}
}
else
{
echo '<label>Data not Found</label>';
}
}
?>
Try with this
sql_query = "SELECT * FROM `countries`,`countries2` WHERE " . $condition;
i have noticed that you are making a search directory or something like this.
you may use FULLTEXT SEARCH with operators and Stemming
after this your query will look like this.
implement as per your requirement ;).
(SELECT * FROM table1 WHERE(col1,col2,clo3) AGAINST(."$search".) IN NATURAL LANGUAGE MODE)
Apologies if I have the terminology wrong.
I have a for loop in php which operates a mysql query...
for ($i = 0; $i <count($user_id_pc); $i++)
{
$query2 = " SELECT job_title, job_info FROM job_description WHERE postcode_ss = '$user_id_pc[$i]'";
$job_data = mysqli_query($dbc, $query2);
$job_results = array();
while ($row = mysqli_fetch_array($job_data))
{
array_push($job_results, $row);
}
}
The results that are given when I insert a...
print_r ($job_results);
On screen -> Array()
If I change the query from $user_id_pc[$i] to $user_id_pc[14] for example I receive one set of results.
If I include this code after the query and inside the for loop
echo $i;
echo $user_id_pc[$i] . "<br>";
I receive the number the counter $i is on followed by the data inside the array for that counter position.
I am not sure why the array $job_results is empty from the query using the counter $i but not if I enter the number manually?
Is it a special character I need to escape?
The full code
<?php
print_r ($user_id_pc);
//Select all columns to see if user has a profile
$query = "SELECT * FROM user_profile WHERE user_id = '" . $_SESSION['user_id'] . "'";
//If the user has an empty profile direct them to the home page
$data = mysqli_query($dbc, $query);
if (mysqli_num_rows($data) == 0)
{
echo '<br><div class="alert alert-warning" role="alert"><h3>Your appear not to be logged on please visit the home page to log on or register. <em>Thank you.</em></h3></div>';
}
//Select data from user and asign them to variables
else
{
$data = mysqli_query($dbc, $query);
if (mysqli_num_rows($data) == 1)
{
$row = mysqli_fetch_array($data);
$cw_job_name = $row['job_description'];
$cw_rate = $row['hourly_rate'];
$job_mileage = $row['mileage'];
$job_postcode = $row['postcode'];
$response_id = $row['user_profile_id'];
}
}
for ($i = 0; $i <count($user_id_pc); $i++)
{
$query2 = " SELECT job_title, job_info FROM job_description WHERE postcode_ss = '{$user_id_pc[$i]}'";
$job_data = mysqli_query($dbc, $query2);
$job_results = array();
while ($row = mysqli_fetch_array($job_data))
{
array_push($job_results, $row);
}
echo $i;
?>
<br>
<?php
}
print ($query2);
print $user_id_pc[$i];
?>
This is primarily a syntax error, the correct syntax should be:
$query2 = " SELECT job_title, job_info FROM job_description WHERE postcode_ss = '{$user_id_pc[$i]}'";
Note that this is correct syntax but still wrong!! For two reasons the first is that it's almost always better (faster, more efficient, takes less resources) to do a join or a subquery or a simple IN(array) type query rather than to loop and query multiple times.
The second issue is that passing parameters in this manner leave your vulnerable to sql injection. You should use prepared statements.
The correct way
if(count($user_id_pc)) {
$stmt = mysqli_stmt_prepare(" SELECT job_title, job_info FROM job_description WHERE postcode_ss = ?");
mysqli_stmt_bind_param($stmt, "s", "'" . implode("','",$user_id_pc) . "'");
mysqli_stmt_execute($stmt);
}
Note that the for loop has been replaced by a simple if
You have to check the query variable, instead of:
$query2 = " SELECT job_title, job_info FROM job_description WHERE postcode_ss = '$user_id_pc[$i]'"
have you tried this:
$query2 = " SELECT job_title, job_info FROM job_description WHERE postcode_ss = '" . $user_id_pc[$i] . "' ";
And another thing, try something different like this:
while ($row = mysqli_fetch_array($job_data))
{
$job_results[] = array("job_title" => $row["job_title"], "job_info" => $row["job_info");
}
Then try to print the values.
Sorry but I like foreach(), so your working code is:
<?php
// To store the result
$job_results = [];
foreach($user_id_pc as $id ){
// selecting matching rows
$query2 ="SELECT job_title, job_info FROM job_description WHERE postcode_ss = '".$id."'";
$job_data = mysqli_query($dbc, $query2);
// checking if query fetch any result
if(mysqli_num_rows($job_data)){
// fetching the result
while ($row = mysqli_fetch_array($job_data)){
// storing resulting row
$job_results[] = $row;
}
}
}
// to print the result
var_dump($job_results);
I'm trying to run an alternate query if the initial query fails (it does because the id I'm searching for in this instance only exists in one of the databases being joined) using an if statement and I've constructed it like so:
<?php
require_once($_SERVER['DOCUMENT_ROOT'].'/php/link_costreport_2013.php');
$id = $_GET['id']; //ID # For page/query
if($query = $link->prepare("SELECT locale.id, locale.provider_num, locale.provider_name, locale.state, locale.city,
finstat_ca.coh_and_banks, finstat_ca.temp_investments, finstat_ca.notes_receivable, finstat_ca.accounts_receivable, finstat_ca.other_receivables,
finstat_ca.afun_and_ar, finstat_ca.inventory, finstat_ca.prepaid_expenses, (finstat_ca.other_cur_assets + finstat_ca.due_from_other_funds) as other_cur_assets, finstat_ca.total_current_assets,
finstat_fa.total_fixed_assets,
finstat_olta.investments, (finstat_olta.dep_on_leases + finstat_olta.due_from_owners_officers + finstat_olta.other_assets) as all_olta, finstat_olta.total_other_assets, finstat_olta.end_assets,
finstat_cl.accounts_payable, finstat_cl.salaries_wages_fees_payable, finstat_cl.payroll_taxes_payable, finstat_cl.notes_loans_payable, finstat_cl.deferred_income, finstat_cl.total_current_liabilities,
(finstat_cl.total_current_liabilities - (finstat_cl.accounts_payable + finstat_cl.salaries_wages_fees_payable + finstat_cl.payroll_taxes_payable + finstat_cl.notes_loans_payable + finstat_cl.deferred_income)) as all_other_cl,
finstat_ltl.mortgage_payable, finstat_ltl.notes_payable, finstat_ltl.unsecured_loans, finstat_ltl.other_long_term_liabilities, finstat_ltl.total_long_term_liabilities,
finstat_talfb.total_fund_balance, finstat_talfb.total_lia_plus_fb
FROM `locale`
INNER JOIN `finstat_ca`
ON locale.id = finstat_ca.id
INNER JOIN `finstat_fa`
ON locale.id = finstat_fa.id
INNER JOIN `finstat_olta`
ON locale.id = finstat_olta.id
INNER JOIN `finstat_cl`
ON locale.id = finstat_cl.id
INNER JOIN `finstat_ltl`
ON locale.id = finstat_ltl.id
INNER JOIN `finstat_talfb`
ON locale.id = finstat_talfb.id
WHERE locale.id = :id
LIMIT 1")){
} else {
$query = $link->prepare("SELECT id, provider_num, provider_name, state, city
FROM `locale`
WHERE id = :id
LIMIT 1");
}
$query->bindParam(':id', $id);
$query->execute();
$results = $query->fetch(PDO::FETCH_ASSOC);
echo json_encode($results);
Basically it defaults to the single table where the ID does exist and only pulls a couple fields as opposed to the large statement above it. My only issue is that my code here is not working. My JSON only says false when I echo it. It obviously should not.
Is there an error in my code here?
Thanks in advance
:edit: I should note that when I enter an ID that exists in all the tables joined, the correct result (json) is displayed on the page.
I believe the problem is that even if ID does not exist in the first query, the $query variable still has a proper query in it and there is nothing false about it. That's not what you should be if-testing.
I think you should be testing $results.
This shows you the logic.
<?php
require_once($_SERVER['DOCUMENT_ROOT'] . '/php/link_costreport_2013.php');
//ID # For page/query
$id = $_GET['id'];
$sql_1 = "SQL CODE FOR QUERY 1";
$sql_2 = "SQL CODE FOR QUERY 2";
$query = $link->prepare($sql_1);
$query->bindParam(':id', $id);
$query->execute();
$results = $query->fetch(PDO::FETCH_ASSOC);
if (!$results)
{
$query = $link->prepare($sql_2);
$query->bindParam(':id', $id);
$query->execute();
$results = $query->fetch(PDO::FETCH_ASSOC);
}
echo json_encode($results);
However as you can see there are a few lines of code that are repeated inside the if-statement that very similar to code that was just before the if-statement. Perhaps with a loop that loops twice but breaks out if $results is not false would be neater.
<?php
require_once($_SERVER['DOCUMENT_ROOT'] . '/php/link_costreport_2013.php');
//ID # For page/query
$id = $_GET['id'];
$sql[] = "SQL CODE FOR QUERY 1";
$sql[] = "SQL CODE FOR QUERY 2";
foreach ($sql as $sql_query)
{
$query = $link->prepare($sql_query);
$query->bindParam(':id', $id);
$query->execute();
$results = $query->fetch(PDO::FETCH_ASSOC);
if ($results)
{
break;
}
}
echo json_encode($results);
The world is your oyster.
i have a query
$result = mysql_query("SELECT * FROM comprofiler WHERE cb_playstationgames LIKE '%FIFA%' ORDER BY id ASC");
in that query there is a user_id wich i need to perform a query on another table.
$gebruikerid = mysql_query("SELECT * FROM users where id LIKE '".$result['user_id']."'");
Now i want to use that value in a while loop
echo "<table><tr><th width=\"300\" align=\"left\" >Avatar</th><th width=\"300\" align=\"left\">Naam</th><th width=\"200\" align=\"left\">PSN Naam</th></tr>";
while($row2 = mysql_fetch_array($result))
{
echo "<tr><td><img height=\"50\" width=\"50\" src=\"/images/comprofiler/" . $row2['avatar'] . "\"></td><td>" . $gebruikernaam . "</td><td>" . $row2['cb_psnnaam'] . "</td></tr>";
}
echo "</table>";
I cannot get the query to read the values from the other table based on the id from the first table. Can someone help me?
In your case the best solution it's to use JOIN.
$result = mysql_query("SELECT comprofiler.*, users.* FROM comprofiler INNER JOIN users ON users.id = comprofiler.user_id WHERE cb_playstationgames LIKE '%FIFA%' ORDER BY comprofiler.id ASC");
You may have to manually specify the columns you want to select (after SELECT ) in case that you have same column names in both table and you need both.
$result = mysql_query("SELECT comprofiler.id as id_comprofiler, users.id as id_user, users.avatar ... " );
To answer your question why it does not work:
Second query does not use mysql_fetch_array.
$gebruikerid = mysql_fetch_array($gebruikerid);
It would be more secure & cleaner if you would use PDO or MySQLi prepared statements like:
$db = new \PDO(SEE PDO CUNSTRUCT)
$query1 = $db->prepare('
SELECT *
FROM comprofiler
WHERE cb_playstationgames LIKE :fifa
ORDER BY id ASC
');
$query2 = $db->prepare('
SELECT * FROM users where id LIKE :id
');
$query1->bindValue(':fifa', '%FIFA%', PDO::PARAM_STR);
$query1->execute();
while ($row = $query1->fetch(PDO::FETCH_ASSOC)) {
$query2->bindParam(':id', $row['user_id'], PDO::PARAM_INT);
$query2->execute();
//Holds associative array of second query
$row2 = $query->fetch(PDO::FETCH_ASSOC);
//$row1 holds associative array of first query
}
I have the following code and all of the search functions work except for the title field. So I can search by genre, date, location etc... but not by title. When attempting to search by title nothing is returned at all. Can anyone help me with this?
Also, is there a more efficient way to count all the fields before limiting it for use in pagination later on?
$today = date("Y-m-d");
$query = "SELECT * FROM TABLE_NAME WHERE Date >= '$today'";
$bind = Array();
if ($_GET["Title"] && $_GET["Title"] != "") {
$query .= " and Title like %?%";
$bind['Title'] = $_GET['Title'];
}
if ($_GET["Genre"] && $_GET["Genre"] != "") {
$query .= " and Genre like %?%";
$bind['Genre'] = $_GET['Genre'];
}
if ($_GET["Location"] && $_GET["Location"] != "") {
$query .= " and Location like %?%";
$bind['Location'] = $_GET['Location'];
}
if ($_GET["Date"] && $_GET["Date"] != "") {
$query .= " and Date = %?%";
$bind['Date'] = $_GET['Date'];
}
$stmt = $db->prepare($query);
$stmt->execute($bind);
$rows = $stmt->fetchAll(PDO::FETCH_ASSOC);
$num = count($rows);
$query .= " ORDER BY Date LIMIT $limit, 9";
$stmt = $db->prepare($query);
$stmt->execute($bind);
$rows = $stmt->fetchAll(PDO::FETCH_ASSOC);
Edit: After everyone's help I thought I would post my now revised code for future reference. It turns out the other fields were not working, but instead due to the if statement all this was nested in the code simply wasn't being executed.
$today = date("Y-m-d");
$query = "SELECT * FROM TABLE_NAME WHERE Date >= '$today'";
$countq = "SELECT count(*) FROM TABLE_NAME WHERE Date >= '$today'";
$bind = Array();
if ($_GET["Title"] && $_GET["Title"] != "") {
$query .= " and Title like :title";
$countq .= " and Title like :title";
$bind[':title'] = "%{$_GET['Title']}%";
}
if ($_GET["Genre"] && $_GET["Genre"] != "") {
$query .= " and Genre like :genre";
$countq .= " and Genre like :genre";
$bind[':genre'] = "%{$_GET['Genre']}%";
}
if ($_GET["Location"] && $_GET["Location"] != "") {
$query .= " and Location like :loc";
$countq .= " and Location like :loc";
$bind[':loc'] = "%{$_GET['Location']}%";
}
if ($_GET["Date"] && $_GET["Date"] != "") {
$query .= " and Date = :date";
$countq .= " and Date = :date";
$bind[':date'] = "{$_GET['Date']}";
}
$stmt = $db->prepare($countq);
$stmt->execute($bind);
$rows = $stmt->fetchAll();
$num = count($rows);
$query .= " ORDER BY Date LIMIT $limit, 9";
$stmt = $db->prepare($query);
$stmt->execute($bind);
$rows = $stmt->fetchAll(PDO::FETCH_ASSOC);
all of the search functions work
With the given query it is not true
From PDO tag wiki:
placeholders cannot represent an arbitrary part of the query, but a complete data literal only. Neither part of literal, nor whatever complex expression or a syntax keyword can be substituted with prepared statement.
Prepare FULL literal first: $name = "%$name%"; and then bind it.
As for the "more" efficient method for pagination - yes, oh yes.
With your current way of counting data you don't actually need other queries. as you have ALL the data already and can paginate it as well.
But of course it will pollute all the memory soon. So, if you want to get a count of rows from database, get the very count: run the same query but instead of SELECT * make it "SELECT count(*)
There are not any errors returned, that's why I am so confused
From PDO tag wiki again:
It is essential to set ERRMODE_EXCEPTION as a connection option as it will let PDO throw exceptions on connection errors. And this mode is the only reliable way to handle PDO errors.