I need to print a row from a database, i know how to print columns, but having a hard time printing rows. Can someone tell me how to?
<?php
$query = "SELECT * FROM categorias ";
$result = mysqli_query($conn, $query) or die (mysql_error());
while ($categoria = mysqli_fetch_array($result)) {
echo "<p>" . $categoria ['descricao'] . "</p>";
}
?>
This is how im printing columns
The answer is don't use SELECT * in PHP, it's extremely prone to errors. If you explicitly list the columns in your select statement you can concatenate them into a table in PHP.
Hope this helps.
Use print_r to debug selected data.
Also look for Mysql Fetch Row
Always Use Google
<?php
$query = "SELECT * FROM categorias ";
$result = mysqli_query($conn, $query) or die (mysql_error());
if(mysqli_num_rows($result)>0)
{
while ($categoria = mysqli_fetch_array($result)) {
echo "<p>" . $categoria['descricao'] . "</p>";
}
}
?>
<table><tr><?php
while ($categoria = mysqli_fetch_array($result)) {
echo "<td>" . $categoria ['descricao'] . "</td>";} ?></tr></table>
I use a table, where while the array is true places the values cell by cell in a row, because the loop is working inside the <tr> </tr> creating a new <td> for every record.
Related
I have populated a dropdown menu with my query results using Mysqli
echo '<select>';
echo '<option>Semester</option>';
$q = "SELECT semester_id FROM semOffered";
$result = mysqli_query($dbc, $q);
while($row = mysqli_fetch_array($result)) {
echo '<option>' . $row['semester_id'] . '</option>';
}
echo '</select>';
$dbc is my database connection
Within my semester_id column I have repeating values. I would like to only display one of these values as a representative of the many.
Is this possible?
For instance, I have:
Number
Number
Number
Number
Number
My goal:
Number
try this
array_unique() to remove duplicate elements or values in an array.
echo '<select>';
echo '<option>Semester</option>';
$q = "SELECT semester_id FROM semOffered";
$result = mysqli_query($dbc, $q);
$result = array_unique($result)
while($row = mysqli_fetch_array($result)) {
echo '<option>' . $row['semester_id'] . '</option>';
}
echo '</select>';
In my database I have a one-to-many table relationship where one parent can have many kids. The primary key is the parents email. I query to get the kids
$results1 = mysqli_query($con,"
SELECT directory.email
, dirKids.kname
, dirKids.kbirthday
FROM directory
JOIN dirKids
ON '$row[email]' = dirKids.parent
");
Then I loop through and echo the value to my html page
while($row1 = mysqli_fetch_array($results1)) {
if (!empty($row1["kname"])) {
echo "<tr><td>". $row1["kname"] ."</td><td>".
$row1["kbirthday"]."</td></tr>";
}
}
The problem I am having is that only one parent has kids in my database, but it will print the kids name and birthday 10 times because there are 10 people in my database. How can I get it to only print the child's name and birthday once?
My full code is listed below:
<?php
$con = mysqli_connect("localhost", "username", "password", "db");
// Check connection
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$results = mysqli_query($con,"SELECT directory.id, directory.fname, directory.lname, directory.address, directory.bdname, directory.birthday, directory.cell, directory.email, directory.sFName, directory.sBirthday, directory.sCell, directory.sEmail FROM directory ORDER BY lname") or die ("couldn't fetch query");
echo "<div class='accordion' id='accordion'>";
// output data of each row
while($row = mysqli_fetch_array($results)) {
$results1 = mysqli_query($con,"SELECT directory.email, dirKids.kname, dirKids.kbirthday FROM directory JOIN dirKids ON '$row[email]' = dirKids.parent");
echo "</table></div>";
if ($row['sFName'] == "" || $row['sFName'] == "undefined") {
echo "<div class='card'><div class='card-header'
id='headingOne'><h5 class='mb-0'><button class='btn btn-link'
type='button' data-toggle='collapse' data-target='#collapse".
$row["id"] ."' aria-expanded='true' aria-controls='collapse".
$row["id"] . "'><h5>".$row["fname"] ."<span id='lnameText'>".
$row["lname"] ."</span></h5></button></h5></div><div
id='collapse". $row["id"] . "' class='collapse'
aria-labelledby='headingOne' data-parent='#accordion'><div
class='card-body'><table id='myUL' class='table'><tr></tr><tr>
<td><h5>Address</h5></td><td>". $row["address"] ."</td></tr>
<tr><td><h5>Birthday</h5></td><td>".$row["birthday"]."</td>
</tr><tr><td><h5>Cell</h5></td><td>". $row["cell"]."</td></tr>
<tr><td><h5>Email</h5></td><td>". $row["email"] ."</td></tr>
</table></div>";
echo "<div class='col-md-6'><h3>Children</h3><table class='table'><tr><th><h5>Name</h5></th><th><h5>Birthday</h5></th>";
while($row1 = mysqli_fetch_array($results1)) {
if (!empty($row1["kname"])) {
echo "<tr><td>". $row1["kname"] ."</td><td>". $row1["kbirthday"]."</td></tr>";
}
}
echo "</table></div></div>";
?>
Since the data needed for the second while loop comes exclusively from the kids table, just build your SELECT statement for that, forget the join and the WHERE statement looks for only the parents email.
The below code goes inside the primary while loop and replaces the
$results1 = mysqli_query($con,"SELECT directory.email, dirKids.kname, dirKids.kbirthday FROM directory JOIN dirKids ON '$row[email]' = dirKids.parent");
with
//Build the select statement
$sql = "SELECT kname, kbirthday FROM dirKids WHERE parent = '" .$row[email] . "'";
//now run the query
$results1 = mysqli_query($con,$sql);
//uncomment the below to see the results
//var_dump(mysqli_fetch_array($results1));
Your query should look like this;
$select = mysqli_query($db, "SELECT * FROM parents_database WHERE parent_name = '$parent_name'");
while ($row = mysqli_fetch_array($select, MYSQLI_ASSOC)) {
// echo kids here..
}
Not sure what do you need. Since you posted 2 different queries.
But 1st one has wrong approach, hope you need to fix that one.
I think you've meant something like:
SELECT directory.email
, dirKids.kname
, dirKids.kbirthday
FROM directory
JOIN dirKids
ON directory.email = dirKids.parent
WHERE directory.email = '$row[email]'
I have had a long road to get to this last question. Everything is my code is working now, but I can't get this last little issue. Right now I have:
$sql = "SELECT phonenumber,email, dataplan AS currentplan, SUM(datamb) AS
value_sum FROM maindata GROUP BY phonenumber, dataplan";
$result = mysql_query($sql);
$row = mysql_fetch_assoc($result);
$val = $row["value_sum"];
$plan = $row["currentplan"];
$remain = $plan - $val;
if (!$result) {
echo "Could not successfully run query ($sql) from DB: " . mysql_error();
exit;
}
if (mysql_num_rows($result) == 0) {
echo "No rows found, nothing to print so am exiting";
exit;
}
It only subtracts the first value as opposed to the values for all. displayed like this:
while ($row = mysql_fetch_assoc($result)){
echo "<tr>";
echo "<td>".$row['phonenumber'] . "</td> ";
echo "<td>".$row['currentplan'] . "</td> ";
echo "<td>".ROUND ($row["value_sum"],2) . "MB</td> ";
echo "<td>".$remain . " MB</td> ";
echo "<td>".$row['email'] . "</td></tr>";
}
So my goal is to subtract all value_sums from all dataplans, but what I have now, gives me the first value for all columns. Thank you!
mysql_fetch_assoc() will always get one row. You can use it in loop, or better use PDO, eg. like this:
$sql = "SELECT phonenumber,email, dataplan AS currentplan, SUM(datamb) AS
value_sum FROM maindata GROUP BY phonenumber, dataplan";
$results = $pdo->query($sql);
You can read about creating PDO connections here http://www.php.net/manual/en/book.pdo.php
i cannot get a row to delete as the id is not going through the url. its a simple error somewhere and i cannot find the solution after having a look around for an hour.
this page contains the information on a table:
<?php
$result = mysql_query("SELECT review, ratings, date, user FROM reviews")
or die(mysql_error()); ;
if (mysql_num_rows($result) == 0) {
echo 'There Arent Any Reviews Yet';
} else {
echo "<table border='0'><table width=100% border='6'><tr><th>Comments/Thoughts</th><th>Ratings</th><th>Date</th><th>User</th><th>Delete</th></tr>";
while($info = mysql_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $info['review']. "</td>";
echo "<td>" . $info['ratings']. " Stars</td>";
echo "<td>" . $info['date']. "</td>";
echo "<td>" . $info['user']. "</td>";
echo "<td>" . " <a href='deletereview.php?review_id=" . $info['review_id'] . "'>Delete</a> </td>";
echo "</tr>";
}
}
echo "</table>";
?>
it goes to deletereview.php which carries out the delete function:
<?php
session_start();
require_once '../includes/db.php';
$id = $_GET['review_id'];
$info = "DELETE FROM reviews WHERE review_id = '$id'";
mysql_query($info) or die ("Error: ".mysql_error());
echo "<h2>Review Deleted</h2>";
?>
any ideas guys?
You're not selecting the review_id in the query, so $info["review_id"] is always null.
Aside from the other answers, I'll say this:
Your database will get jacked if you do not sanitize your variables.
For instance, what happens if I pass review_id=' OR '1'='1?
DELETE FROM reviews WHERE review_id = '' OR '1'='1'
This query will delete everything in reviews.
mysql_real_escape_string() your $_GET and $_POST variables before using them in your MySQL.
You forgot to select the review_id.
$result = mysql_query("SELECT review_id, review, ratings, date, user FROM reviews")
You're not selecting review_id from the database but you use $info['review_id'] to set the ID on the URL. Just change your first line to:
$result = mysql_query("SELECT review_id, review, ratings, date, user FROM reviews")
Also you must escape the input with mysql_real_escape_string:
$id = mysql_real_escape_string($_GET['review_id']);
You have to select the review_id in the query. But also you have to check for some SQL injection, because with the GET request it's easy to delete all the table records.
I am trying to delete the records from the users table in mysql,
the code goes like this.
if(isset($_GET['id'])) {
//create query to delete the record
$query = "DELETE FROM users WHERE id =" . int($_GET['id']) or die(mysql_error());
//execute query
if($mysqli->query($query)) {
//print number of affected rows
echo $mysqli->affected_rows. " row(s) affected";
}
else {
//print error message
echo "Error in query : $query " . $mysqli->error;
}
}
else {
echo "Could not Execute the Delete query";
}
at the same time i am iterating the records from the users table in the database and it goes like this.
//query to get records
$query = "SELECT * FROM users";
//execute query
if($result = $mysqli->query($query)) {
// see if any rows were returned
if($result->num_rows > 0) {
// if yes then print one after another
echo "<table cellpadding=10 border=1>";
while($row = $result->fetch_array()) {
echo "<tr>";
echo "<td>" .$row[0] . "</td>";
echo "<td>" .$row[1] . "</td>";
echo "<td>" .$row[2] . "</td>";
echo "<td>Delete</td>";
echo "</tr>";
}
echo "</table>";
}
$result->close();
}
the problem is, i am able to get the records from the database and display it in the browser but when i try to delete the record the first condition does not pass i.e if(isset($_GET['id'])) instead it goes to else condition and print the message "Could not Execute the Delete query " , i guess it is not able to fetch the $_GET['id'] so only it refuses to enter the if condition,
P.S :i would appreciate if someone explains me in simple words, i am a newbie to programming, thanks..
You are missing an =:
echo "<td>Delete</td>";
HERE -------------------^
"DELETE FROM users WHERE id =" . int($_GET['id']) or die(mysql_error());
Shouldn't it be intval instead? There's no function int in PHP. There's also (less preferably) the cast to int, like this: (int) $_GET['id']).