$data = mysql_query(" SELECT * FROM user_pokemon_db WHERE user_id = '".$id."' ");
while($rows = mysql_fetch_array($data))
{
$pkmn_id = $rows['pkmn_id'];
$path = mysql_query(" SELECT path FROM pokemons WHERE pk_id = '".$pkmn_id."' ");
$poke = mysql_result($path, 0, "path");
echo $poke;
echo "<br />";
$level = $rows['level'];
echo $level;
echo "<br />";
$exp = $rows['exp'];
echo $exp;
This is my PHP code, its showing an error:
Array to string conversion in C:\wamp\www\slots.php on line 18
Line 18 is this:while($rows = mysql_fetch_array($data))
I haven't used any array?? but this used to work! but suddenly this error started coming??
Check if $id is an array, it seems that it causes this problem.
$id = "'" . implode("', '", $id) . "'";
$data = mysql_query(" SELECT * FROM user_pokemon_db WHERE user_id IN ({$id}) ");
Related
I've been throw so many threads for 4+ hours here and abroad and seem to be missing a simple thing.
I'm trying to have several users upload their 'news' to MYSQL.
Yet I want to display only the 'news' with the logged in username (userpost) attached to the row.
$current is the username for who is logged in, which works.
Example A isn't filtering out rows that don't contain the $current user.
Example B isn't providing any output
So I've tried both A:
$result = mysqli_query($con,"SELECT * FROM images_tbl");
//echo $current . "2" . $current;
while($row = mysqli_fetch_array($result)) {
if ($row['userpost'] = '.$current.') {
$num = 0;
$num = $num + 1;
$pic.$num = $row['images_path'];
$h1 = $row['hlone'];
and B:
$result = mysqli_query($con,"SELECT * FROM images_tbl WHERE (userpost = '.$current.')");
echo $current . "2" . $current;
while($row = mysqli_fetch_array($result)) {
echo $row['hlone'] . " " . $row['images_path'];
echo "<img src=\"" .$row['images_path']. "\">";
}
27, images/08-10-2014-1412752801.jpg(images_path), 2014-10-08, Headline(hlone), Headline2, story, testb(userpost)
Any help would be greatly appreciated.
Add where clause to your query
//in situation A
$result = mysqli_query($con,"SELECT * FROM images_tbl where username='".$current."'");
//username is column name for user you might have to change this
while($row = mysqli_fetch_array($result)) {
echo $row['images_path'];
echo $row['hlone'];
}
In situation B try this
$result = mysqli_query($con,"SELECT * FROM images_tbl WHERE userpost = '".$current."')");
echo $current . "2" . $current;
while($row = mysqli_fetch_array($result)) {
echo $row['hlone'] . " " . $row['images_path'];
echo "<img src=\"" .$row['images_path']. "\">";
}
Why are you trying to filter with PHP.
If you want to filter the 'news' that have not written by current user just use MySQL Where clause:
// For Example A
$result = mysqli_query($con, "SELECT * FROM images_tbl WHERE userpost != '{$current}'");
while($row = mysqli_fetch_array($result)) {
$pic = $row['images_path'];
$h1 = $row['hlone'];
}
// For Example B
$result = mysqli_query($con,"SELECT * FROM images_tbl WHERE userpost = '{$current}')");
echo $current . "2" . $current;
while($row = mysqli_fetch_array($result)) {
echo $row['hlone'] . " " . $row['images_path'];
echo "<img src=\"" .$row['images_path']. "\">";
}
It's easy with MySQL's filtering options. You should do more research about MySQL.
I have a number pages namely
change=1.php
change=2.php
change=3.php
They all have similar coding but leaving 1 or 2 variable values.
And I know its a very bad idea! How can I make a link work like below:
change.php?id=1
change.php?id=2
change.php?id=3
http://oi62.tinypic.com/708gfm.jpg
<?php
include 'connection.php';
session_start();
include 'details.php';
/*$pkmn_id = $_SESSION['pkmn_id'];
$poke = $_SESSION['path'];*/
$data = mysql_query(" SELECT * FROM user_pokemon_db WHERE team = 1 AND user_id = '".$id."' ");
while($rows = mysql_fetch_array($data))
{
$rep_id = $rows[0];
$pkmn_id = $rows['pkmn_id'];
$path = mysql_query(" SELECT * FROM pokemons WHERE pk_id = '".$pkmn_id."' ");
$poke = mysql_result($path, 0, "path");
echo $poke;
echo "<br />";
$level = $rows['level'];
echo $level;
echo "<br />";
$exp = $rows['exp'];
echo $exp;
echo "<br />";
echo "<br />";
}
$data = mysql_query(" SELECT * FROM user_pokemon_db WHERE user_id = '".$id."' AND team = 0");
while($rows = mysql_fetch_assoc($data))
{
$db_id = $rows['id'];
$array[] = $db_id;
$level = $rows['level'];
$array1[] = $level;
$exp = $rows['exp'];
$array2[] = $exp;
$pkmn_id = $rows['pkmn_id'];
$data1 = mysql_query(" SELECT * FROM pokemons WHERE pk_id = '".$pkmn_id."' ");
while($rows = mysql_fetch_assoc($data1))
{
$poke = $rows['path'];
$array3[] = $poke;
}
}
$team = 1;
$_SESSION['team'] = $team;
$_SESSION['rep_id'] = $rep_id;
?>
My PHP code.
You probably want to use GET variables, for which you need to combine all the files into one, named change.php. In this file you need the line $foo = $_GET["id"] which will get the value of the variable "id" in the url change.php?id=1.
if (isset($_GET["id"])) {
$foo = $_GET["id"];
//your code here
}else{
echo 'ERROR!!! No id in URL';
}
You can have several variables in the URL like this: change.php?id=1&a=bar&b=toofoo
You can get current script's file name and parse integer.
__FILE__
gives current script's name. Then,
$myStr = preg_replace('/\.php$/', '', __FILE__);
$result = preg_replace('/change=$/', '', $myStr);
echo $result; // it's your id
I am trying to return a value from my database based on a user input on my form.
When I run the code using a value it works but when I put in the variable it doesn't. I am sure it is something simple, but I just don't get it?
Here is the code that works:
$sql_beam = mysqli_query($link,"SELECT cost_ft FROM Beams WHERE number = '201'");
while($row = mysqli_fetch_array($sql_beam))
{
echo "<p>" . $row['cost_ft'] . "</p>";
echo "<br>";
}
When I change it to this it doesn't:
$beam_num = $_POST['Beam Number'];
$sql_beam = mysqli_query($link,"SELECT cost_ft FROM Beams WHERE number = '$beam_num'");
while($row = mysqli_fetch_array($sql_beam))
{
echo "<p>" . $row['cost_ft'] . "</p>";
echo "<br>";
}
You should never put any variable directly into a query. Google sql injection and how to prevent it.
Here is a simple example:
$sql = "SELECT cost_ft FROM Beams WHERE number = ?";
$stmt = $link->prepare($sql);
$stmt->bind_param("s", $beam_num);
$stmt->execute();
$result = $stmt->get_result();
while($row = $result->fetch_assoc())
{
echo "<p>" . $row['cost_ft'] . "</p>";
echo "<br>";
}
Variables in a query work best when enclosed in curly brackets like this:
$beam_num = $_POST['Beam Number'];
$sql_beam = mysqli_query($link,"SELECT cost_ft FROM Beams WHERE number = '{$beam_num}'");
while($row = mysqli_fetch_array($sql_beam))
{
echo "<p>" . $row['cost_ft'] . "</p>";
echo "<br>";
}
That should solve the problem, if it doesnt then try this:
$sql_beam = mysqli_query($link,"SELECT cost_ft FROM Beams WHERE number = $beam_num");
I had that in one of my codes and it worked.
Change your second line to this:
$sql_beam = mysqli_query($link,"SELECT cost_ft FROM Beams WHERE number = '". $beam_num ."'");
use query for access the $current_rank this value want to access in different query but this value can not access any where in different query so how to access $current_rank......
$query = "select * from menu_master where menu_id =
$row_id and hotel_id='" . $_REQUEST['hotel_id'] . "'";
$result = mysql_query($query)."<br/>";
while($row=mysql_fetch_array($result))
{
$rank = $row['set_rank'];
}
$current_rank = $rank;
//echo $current_id = $row_id."<br/>";
//echo $new_rank =$_REQUEST['set_rank']."<br/>";
$sql = "select * from menu_master where set_rank = '$new_rank ' and hotel_id='".$_REQUEST['hotel_id']."'" ;
// echo $sql."<br/>";
$rs = mysql_query($sql)."<br/>";
while($row = mysql_fetch_array($rs))
{
$menu_id = $row['menu_id'];
$sql="update menu_master
set set_rank=$current_rank where menu_id= $menu_id and hotel_id='".$_REQUEST['hotel_id']."'";
//echo $sql."<br/>";
mysql_query($sql)."<br/>";
}
$sql="update menu_master set menu_name = '" . mysql_real_escape_string($_REQUEST['menu_name']) . "',
menu_name_ar = '" . mysql_real_escape_string($_REQUEST['menu_name_ar']) . "',
is_active = '" . $is_active . "',
set_rank = $new_rank where menu_id = '$current_id' and hotel_id='".$_REQUEST['hotel_id']."'";
//echo $sql."<br/>";
//exit;
mysql_query($sql);
Your current_rank seems to be an array. If you have single value in current_rank, then do not use while loop for it.
Just use $row=mysql_fetch_array($result);
$current_rank = $row['set_rank'];
Also you have commented out this line.
//echo $new_rank =$_REQUEST['set_rank']."";
So you have no value for $new_rank
I just can't figure out why i get the error message, I have tried removing the'' and the()
I have run the script in phpmyadmin and it says the problem with my syntax is at $result = ("SELECT * FROM 'test_prefixCatagory' ORDER by 'Cat'");
$result = ("SELECT * FROM 'test_prefixCatagory' ORDER by 'Cat'");
while($row = mysql_fetch_array($result))
$sCat = ($row['Cat']);
$sCatID = ($row['CatID']);
{
echo "<table>";
echo "<tr valign='top'><td><b><a href='#".$sCat."'>".$sCat."</a></b><br>";
// column 1 categories
$result2 = ("SELECT * FROM `test_prefixSubCat` WHERE `CatID`=$sCatID");
// sub-cats
while($row2 = mysql_fetch_array($result2))
{
$sSub = ($row2['CatID']);
$sSubID = ($row2['SubID']);
echo "<dd><a href='#'>".$sSub."</a><br>";
}
echo "<br></td></tr>";
echo "</table>";
}
Do anyone have an idea?
Try this :
<?php
$result = mysql_query("SELECT * FROM `test_prefixCatagory ORDER by `Cat`");
while ($row = mysql_fetch_array($result)) {
$sCat = $row['Cat'];
$sCatID = $row['CatID'];
echo "<table>";
echo "<tr valign='top'><td><b><a href='#" . $sCat . "'>" . $sCat . "</a></b><br>";
// column 1 categories
$result2 = mysql_query("SELECT * FROM `test_prefixSubCat` WHERE `CatID`='".$sCatID. "'");
// sub-cats
while ($row2 = mysql_fetch_array($result2)) {
$sSub = $row2['CatID'];
$sSubID = $row2['SubID'];
echo "<dd><a href='#'>" . $sSub . "</a><br>";
}
echo "<br></td></tr>";
echo "</table>";
}
?>
$result = ("SELECT * FROM `test_prefixCatagory` ORDER by `Cat`");
Not only do you need to add mysql_query but you also need to remove the single quotes from the table name and field name. You can use backticks if you wish but not single quotes around table names.
$result = mysql_query("SELECT * FROM `test_prefixCatagory` ORDER by `Cat`");
// other query:
$result2 = mysql_query("SELECT * FROM `test_prefixSubCat` WHERE `CatID`=$sCatID");
When debugging MySQL problems, use mysql_error() to see a description of the problem.