I'm trying to select all data from all tables that belong to a specific database with MySQLi.
So, I am looking for something like SELECT * FROM *
I have the following (awesome) code, but it isn't working. Can someone enlighten me?
<h1>All posts</h1>
<?php
$servername = "localhost";
$username = "Masa";
$password = "passwdhere";
$dbname = "forums";
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT * FROM *";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
echo '<h3>' . $row["title"] . '</h3><br><br><p>' . $row["text"] . '</p><br> <h4>Answers</h4><br><p>' . $row["answer"] .'</p><br> <form method = "post"><label for = "answer">Post your answer</label><br><textarea class = "answer" id = "answer" name = "answer" requred maxlength="10000" ></textarea><br><input type = "submit" class = "submit" name = "submit" id = "submit" value = "Answer"></form><br> <hr style="width: 650px; max-width: 100%; color: #eee;"><br><br> ';//I know, don't print HTML tags :3
}
} else {
echo "No posts yet! :c";
}
$conn->close();
?>
You want to use show tables; - that being said, I'd recommend reading the MySQL documentation or a tutorial this is really basic stuff. I'm sure just searching on SO would have given you the answer too.
Related
I'm trying to show all the platforms in the "area" and "block_number" which is passed to my page in the url. However when I run the page with the "area" and "block_number" in the url I am not getting any rows. I only get the else statement Sorry No Results. Can someone please show me what I'm doing wrong?
<?
//// Get items from url
$area1 = $_GET[area];
$block_number1 = $_GET[block_number];
$servername = "localhost";
$username = "";
$password = "";
$dbname = "";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = 'SELECT * from platform_locations where area = $area1 and block_number = $block_number1';
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc())
{
echo '<div class="col-lg-4 col-md-4 col-sm-6 wow bounceInUp" data-wow-duration="1s" data-wow-delay="0.4s">
<div class="feature-1 box-shadow">
<img src="img/platform_icon.png" alt="">
<h3 class="feature__title">'. $row['area'] .' '. $row['block_number'] .'</h3>
<p class="feature__text">
Structure Name: '. $row['structure_name'] .'<br>
Attended: '. $row['attended'] .'<br>
More Information
<br>
<font color="Tomato">'. $row['remove_date'] .'</p>
</div></div>';
}
}
else{
echo '<br>Sorry No Results';
}
?>
There was a space in between the equals symbol and $area1 that was making it return 0 rows. I do believe that the other comments were also correct and I will be researching ways to make my code more secure as #Barmar suggested.
area = $area1
I changed area =$area1 and it started working as expected.
Can I please have some help with a problem I'm having updating a mysql database with PHP.
I'm sorry to ask a question that has been asked a lot of times before, it's just driving me a bit nuts, and I've looked through similar questions but the answers don't seem to help with my problem.
I'm using two files, an admin page (admin.php) to edit content with, and an update file that is meant to update the database when the submit button is pressed.
Everything seems to be working fine, the values are being posted to the update.php page (I can see them when I echo them out) but it wont update the database.
If anyone can please point me in the right direction or tell me what I'm doing wrong I'd be very grateful!
Thank you very much:)
This is my admin.php page;
<head>
<?php
/*
Check to see if the page id has been set in the url.
If it has, set it as the $pageid variable,
If it hasn't, set the $pageid variable to 1 (Home page)
*/
if (isset($_GET['pageid'])) {
$pageid = $_GET['pageid'];
}
else {
$pageid = '1';
}
//Database connection variables
$servername = "localhost";
$username = "root";
$password = "";
$database = "cms";
// Create connection
$conn = new mysqli($servername, $username, $password, $database);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
//Get information from the database
$sql = "SELECT title, sub_title, tab1, tab2, tab3, content FROM data WHERE id='$pageid'";
$result = $conn ->query($sql);
if ($result->num_rows > 0)
{
while($row = $result->fetch_assoc()) {
$conn->close();
//Store database information in variables to display in the form
$title = $row["title"];
$sub_title = $row["sub_title"];
$tab1 = $row["tab1"];
$tab2 = $row["tab2"];
$tab3 = $row["tab3"];
$content = $row["content"];
}
} else {
echo "0 results";
}
?>
</head>
<body>
//basic navigation
Page 1 | Page 2 | Page 3
<form action="update.php" method="post" name="adminform">
<input type="hidden" name="pageid" value="<?php echo "$pageid";?>">
NAME:<br>
<input type="text" name="title" value="<?php echo $title;?>"><br><br>
EMAIL:<br>
<input type="text" name="sub_title" value="<?php echo $sub_title;?>"><br><br>
CONTENT:<br>
<input type="text" name="tab1" value="<?php echo $tab1;?>"><br><br>
CONTENT:<br>
<input type="text" name="tab2" value="<?php echo $tab2;?>"><br><br>
CONTENT:<br>
<input type="text" name="tab3" value="<?php echo $tab3;?>"><br><br>
CONTENT:<br>
<textarea rows="4" cols="50" name="content">
<?php echo $content;?>
</textarea>
<br><br>
<input type="submit">
</form>
</body>
And this is the update.php page;
<?php
/*Values passed from the admin form, to be used as update variables*/
if (isset($_POST['adminform']))
{
$pageid = $_POST["pageid"];
$titleu = $_POST["title"];
$sub_titleu = $_POST["sub_title"];
$tab1u = $_POST["tab1"];
$tab2u = $_POST["tab2"];
$tab3u = $_POST["tab3"];
$contentu = $_POST["content"];
}
?>
<?php
if(isset($_POST['adminform']))
{
// Create connection
$conn = new mysqli($servername, $username, $password, $database);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
//Update the database
$sql = "UPDATE data SET title='$titleu', sub_title='$sub_titleu', tab1='$tab1u', tab2='$tab2u', tab3='$tab3u', content='$contentu' WHERE id =='$pageid'";
$result = $conn ->query($sql);
$conn->close();
}
?>
You're using == instead of = on the where clause.
On the other hand, don't pass user values to the query without validation and sanitization if you don't want to be vulnerable to sql injection attacks.
$sql = "UPDATE data SET title='" . $conn->real_escape_string($titleu) . "', sub_title='" . $conn->real_escape_string($sub_titleu) . "', tab1='" . $conn->real_escape_string($tab1u) . "', tab2='" . $conn->real_escape_string($tab2u) . "', tab3='" . $conn->real_escape_string($tab3u) . "', content='" . $conn->real_escape_string($contentu) . "' WHERE id = " . (int)$pageid;
This will work, but is not very elegant solution. You may use prepared statements instead, to pass the correct types and prevent sql injection.
Check your DB Connections and test whether you are connected to DB or not.
Change your query as below
$sql = "UPDATE data SET title='".$titleu."', sub_title='".$sub_titleu."', tab1='".$tab1u."', tab2='".$tab2u."', tab3='".$tab3u."', content='".$contentu."' WHERE id ='$pageid'";
I'm making a page where you have to enter a text in a textbox and the click send, another page will save it.
Also, on the first page, the text that was stored previously in the database, has to load. This is the code that i've got:
<?php
$databaseid = 3;
$servername = "jog4fun.be.mysql";
$username = "jog4fun_be";
$password = "****";
$dbname = "jog4fun_be";
$gettitel1 = null;
$gettext1 = null;
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT Id,Titel,Tekst FROM Teksten";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
if ($row["Id"]== $databaseid ){
$gettitel1 = $row["Titel"];
$gettext1 = $row["Tekst"];
}
}
} else {
echo "0 results";
}
$conn->close();
$gettitel1 = strip_tags($gettitel1, '<br>');
$link1 = '<textarea id = "klein" rows="4" cols="50" name="titel3" form="usrform">' . $gettitel1 . '</textarea>';
$link2 = '<textarea id = "groot" rows="4" cols="50" name="text3" form="usrform">' . $gettext1 . '</textarea>';
echo $link1;
echo $link2;
?>
The problem is that it sends the text from textbox with name text3 as text1 with the post function. Can someone figure out what's wrong with it? I've been tying for an hour and i did not find it.
Thanks for your time and help,
Jonas
Simply here:
while($row = $result->fetch_assoc()) {
if ($row["Id"]== $databaseid ){
$gettitel1 = $row["Titel"];
$gettext1 = $row["Tekst"];
}
you overwrite the value of the two variables each time you iterate. So after this block of code you will keep stored the last values returned from the db.
YOu should add to your query a WHERE clause to identify the one and only record you need so you will fetch only the relevant data. Example:
$sql = "SELECT Id,Titel,Tekst FROM Teksten WHERE Id='1'";
i am trying display a the rows in my database table using the array of ids gotten from another table. i want it to display the first row which is $rowsfriend. and display the second row which is rows .......... but it only displays $rowsfriend
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "ochat";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT * FROM friends where friend1='".($_POST[id])."'";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
$rowsfriend = $row["friend2"];
echo $rows;
}
}
$sqll = "SELECT * FROM users WHERE id IN ($rowssfriend)";
$resultt = $conn->query($sqll);
if ($resultt->num_rows > 0) {
while($roww = $resultt->fetch_assoc()) {
$rowsss = $row["username"];
echo $rowss;
}
}
else {
?>
<h1>Register</h1>
<form action="selectfriends.php" method="post">
id:<br />
<input type="text" name="id" value="" />
<br /><br />
<input type="submit" value="enter" />
</form>
<?php
}
?>
Instead of two queries, you can write this nested query.
$sqll = "SELECT * FROM USERS WHERE ID IN (SELECT friend2 FROM friends WHERE friend1='".$_POST[$id]."')";
$resultt = $conn->query($sqll);
if ($resultt->num_rows > 0)
{
while($roww = $resultt->fetch_assoc())
{
$rowsss = $row["username"];
echo $rowss;
}
}
Hope this solve your problem .
Please try this version of code instead, if you might:
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "ochat";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT users.* FROM users WHERE users.id IN (SELECT friend2 FROM friends where friend1='".($_POST[id])."')";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
$rows = $row["username"];
echo $rows;
}
}
else {
?>
<h1>Register</h1>
<form action="selectfriends.php" method="post">
id:<br />
<input type="text" name="id" value="" />
<br /><br />
<input type="submit" value="enter" />
</form>
<?php
}
?>
As far as I understood the code well, the problem was with the variable name typo as #Admieus wrote but also in the fact that in each iteration of the first loop variable $rowsfriend got overriden with a new value so after the end of the loop, $rowsfriend contained the last id from the result of the first query $sql.
The above version makes only one query using subquery in it to get directly usernames who are friends of friend1 given in $_POST[$id].
I hope it helps.
There are typos in the second loop.
You assign the value to $rowsss and try to echo from $rowss notice the difference.
Also, you assign the fetch_assoc result to $roww and then try to call it again with $row.
$sqll = "SELECT * FROM users WHERE id IN ($rowsfriend)";
$resultt = $conn->query($sqll);
if ($resultt->num_rows > 0) {
while($roww = $resultt->fetch_assoc()) {
$rowsss = $roww["username"];
echo $rowsss;
}
}
Point of improvement is: check your variable names, make names that are easy to understand and hard to mix up.
For instance, the variable containing the sql query should not be named $sql and to make it worse a second query shoul not be named sqll. Instead use names that imply what you are doing.
$querySelectFriendsFrom = "SELECT * FROM users WHERE id IN ($friendId)";
Don't take this as a hard rule, it's more of a tip to prevent silly mistakes.
Update: there was also a type in the query referring to rowssfriend instead of rowsfriend. Fixed above.
I'm trying to bring data from MYSQL Database called ebms_db. The table is events and the fields are Event_ID and Event_Name.
The code I'm using currently to show the Event_Name only in the dropdown list is:
<select name="mySelect">
<?php
include 'db.php';
$sql = "SELECT Event_Name FROM events";
$result = mysql_query($sql);
echo "<select name='Event_Name'>";
while ($r = mysql_fetch_array($result)) {
echo '<option value="'.$row["Event_Name"].'">'.$row["Event_Name"].'</option>';
}
echo "</select>";
?>
<input type = "submit" name="Search" value="Search">
</select>
Db.Php looks like this...
$servername = "localhost";
$username = "test";
$password = "test";
$dbname = "ebms_db";
$conn = new mysqli($servername, "test", "test", $dbname);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
echo "Error";
}
What am I doing wrong?
The output shows a combo box with
- '.$row["Events_Name"].'
inside it.
while ($row = mysql_fetch_array($result)) {
^^^^ here
echo '<option value="'.$row["Event_Name"].'">'.$row["Event_Name"].'</option>';
}
<?php
include 'db.php';
// you just fetching here (Event_Name) take all the values from the database or
$sql = "SELECT * FROM events";
$result = mysql_query($sql);
echo "<select name='Event_Name'>";
while ($row = mysql_fetch_array($result)) {
echo '<option value="'.$row["Event_Name"].'">'.$row["Event_Name"].'</option>';
}
echo "</select>";
?>
<input type = "submit" name="Search" value="Search">
Why are you taking two selects? I just removed one select just beginning above the php tags.
MYSQL is deprecated, you should really use something else, if and/or when they remove it your website will be defunct.