PHP - Php Array Error In Mysql Query

I have tried many ways of storing an array of data that comes from a multiple selection form into a mysql table, including serialization.  I am currently trying to do it by forming a string, and using the string in the insert query, but keep getting errors, or just nothing happening.  I am not sure what I am doing wrong. 

Code: [Select]
$categoryString = array();

if ($categoryArray){
foreach ($categoryArray as $category){
$categoryString[] =  $category.'<br />';}
}
 $categoryquery= "INSERT INTO categoryname VALUES('$categoryString')";

mysql_query($categoryquery);
Thank you for your time as always.

Hello Guys,

I have a question

I have the following query


$result3 = mysql_query("SELECT * FROM table1 WHERE ad_id='$id2'")
or die(mysql_error());  


$row3 = mysql_fetch_array( $result3 );

// Grab all the var
$features = $row3['features'];


I am turning it into an array (comma separated)


$feature2 = explode(",", $features);
print_r($feature2);
)

The result is like so


Array ( [0] => 5 [1] => 9 [2] => 13




I want to query the features for just the ids (F_name) are for the features . This query will show all.. I would like to just display the f_name values from the array query.


// build and execute the query
$sql = "SELECT * FROM features";
$result = mysql_query($sql);
 
// iterate through the results
while ($row = mysql_fetch_array($result))
{
  echo $row['f_name'];
  echo "<br />";
}



Please Advise..

Thanks, Dan

The page that I'm currently working on has a form which allows, through a series of check boxes and text entry fields, a dynamic mysql query to be created. The php code constructs the query based on which fields the user updated in order to find a specific user in the database. Everything was working fine until I started to only display part of the result set. In order to display a message similar to "Displaying 12-25 of 65 results." I ran the query twice. Once to get the total number of results, and once to only get the specified set.

The code for the section producing the error is included below. The elseif statement works as desired, as does the initial query of the if statement.
if (isset($sql) && $sql !=""){
$sql = "SELECT DISTINCT users.StudentID FROM users lEFT JOIN events ON users.StudentID=events.StudentID WHERE" . $sql;
$stmt = $mysqli->prepare($sql);
call_user_func_array(array(&$stmt, 'bind_param'), $array_of_params);
$stmt->execute();
$stmt->store_result();
$total= $stmt->num_rows;
$stmt->close();
$sql = "SELECT DISTINCT users.StudentID,users.FirstName,users.LastName,users.Email,users.Phone,users.Texting,users.Admin,users.HDHS,users.Sex,users.Confirmation FROM users lEFT JOIN events ON users.StudentID=events.StudentID WHERE" . $sql . " ORDER BY LastName,FirstName LIMIT " . $_GET['start'] . ",12";
$stmt = $mysqli->prepare($sql);
call_user_func_array(array(&$stmt, 'bind_param'), $array_of_params);
$stmt->bind_result($row['StudentID'],$row['FirstName'],$row['LastName'],$row['Email'],$row['Phone'],$row['Texting'],$row['Admin'],$row['HDHS'],$row['Gender'],$row['Confirmation']);

}
elseif(isset($_GET['submit']) || isset($_GET['earlier']) || isset($_GET['later'])){
$stmt = $mysqli->prepare("SELECT StudentID,FirstName,LastName,Email,Phone,Texting,Admin,HDHS,Sex,Confirmation FROM users ORDER BY LastName,FirstName");
$stmt->bind_result($row['StudentID'],$row['FirstName'],$row['LastName'],$row['Email'],$row['Phone'],$row['Texting'],$row['Admin'],$row['HDHS'],$row['Gender'],$row['Confirmation']);
$stmt->execute();
$stmt->store_result();
$total= $stmt->num_rows;
$stmt = $mysqli->prepare("SELECT StudentID,FirstName,LastName,Email,Phone,Texting,Admin,HDHS,Sex,Confirmation FROM users ORDER BY LastName,FirstName LIMIT " . $_GET['start'] . ",12");
$stmt->bind_result($row['StudentID'],$row['FirstName'],$row['LastName'],$row['Email'],$row['Phone'],$row['Texting'],$row['Admin'],$row['HDHS'],$row['Gender'],$row['Confirmation']);
$stmt->execute();
}


If anyone would be willing to help me out here it would be greatly appreciated. If you need more information let me know. Thank you!

Hi all,

I have the following code that queries a MySQL database and outputs results on-screen.

Code: [Select]
$result= mysql_query
        ("
        SELECT email,firstname,lastname FROM tablename WHERE ID IN('".$IDs."')
        ");
        if (!$result) {die('Error: Could not search database. ' );}
   
        while($row = mysql_fetch_assoc($result))
        {
            echo $row['firstname'],"&nbsp;",$row['lastname'],"<br />";
            $emails[]=$row['email'];
        }
        var_dump($emails); var_dump($IDs);

However, the two var_dump's output the following (in the case of 2 results for the query):

array(6) { => string(14) "email1@email.com" [1]=> string(23) "email2@email.com" [2]=> string(14) "email1@email.com" [3]=> string(23) "email2@email.com" [4]=> string(14) "email1@email.com" [5]=> string(23) "email2@email.com" } string(67) "9543219aa68ffa1d434dc8530f23fe48','d4384b2b493867706b0bcedda012ed48"

($IDs is an imploded array)

Even though the MySQL table only contains one email per ID, both emails are listed 3 times in $emails, instead of just once. Can anyone see why this is? I'm stuck.

Thanks!

I have these two tables... schedule (gameid, homeid, awayid, weekno, seasonno) teams (teamid, location, nickname) This mysql query below gets me schedule info for ALL 32 teams in an array...

$sql = "SELECT
h.nickname AS home,
a.nickname AS away,
h.teamid AS homeid,
a.teamid AS awayid,
s.weekno
FROM schedule s
INNER JOIN teams h ON s.homeid = h.teamid
LEFT JOIN teams a ON s.awayid = a.teamid
WHERE s.seasonno =2014";
$schedule= mysqli_query($connection, $sql);
if (!$schedule) {
die("Database query failed: " . mysqli_error($connection));
} else { 
// Placeholder for data
$data = array();
while($row = mysqli_fetch_assoc($schedule)) {
if ($row['away'] == "") {$row['away']="BYE";}
$data[$row['homeid']][$row['weekno']] = $row['away'];
$data[$row['awayid']][$row['weekno']] = '@ '.$row['home'];
}
}

However, I only want to get info for one specific team, which is stored in the $teamid variable. This should be very easy, right? I have tried multiple things, including this one below (where I added an AND statement of "AND (h.teamid=$teamid OR a.teamid=$teamid)"), but this one still outputs too much...

$sql = "SELECT
h.nickname AS home,
a.nickname AS away,
h.teamid AS homeid,
a.teamid AS awayid,
s.weekno
FROM schedule s
INNER JOIN teams h ON s.homeid = h.teamid
LEFT JOIN teams a ON s.awayid = a.teamid
WHERE s.seasonno =2014
AND (h.teamid=$teamid OR a.teamid=$teamid)";
$schedule= mysqli_query($connection, $sql);
if (!$schedule) {
die("Database query failed: " . mysqli_error($connection));
} else { 
// Placeholder for data
$data = array();
while($row = mysqli_fetch_assoc($schedule)) {
if ($row['away'] == "") {$row['away']="BYE";}
$data[$row['homeid']][$row['weekno']] = $row['away'];
$data[$row['awayid']][$row['weekno']] = '@ '.$row['home'];
}
}

Below is the array that the above outputs. In a nutshell, all I want is that 1st array ([1]) which has, in this example, the Eagles full schedule. It's not giving me too much else and I guess I could live with it and just ignore the other stuff, but I'd rather be as efficient as possible and only get what I need...

Array
(
[1] => Array
(
[1] => Jaguars
[2] => @ Colts
[3] => Redskins
[4] => @ 49ers
[5] => Rams
[6] => Giants
[7] => BYE
[8] => @ Cardinals
[9] => @ Texans
[10] => Panthers
[11] => @ Packers
[12] => Titans
[13] => @ Cowboys
[14] => Seahawks
[15] => Cowboys
[16] => @ Redskins
[17] => @ Giants
)

[27] => Array
(
[1] => @ Eagles
)

[28] => Array
(
[2] => Eagles
)

[4] => Array
(
[3] => @ Eagles
[16] => Eagles
)

[14] => Array
(
[4] => Eagles
)

[15] => Array
(
[5] => @ Eagles
)

[3] => Array
(
[6] => @ Eagles
[17] => Eagles
)

[] => Array
(
[7] => @ Eagles
)

[16] => Array
(
[8] => Eagles
)

[25] => Array
(
[9] => Eagles
)

[11] => Array
(
[10] => @ Eagles
)

[7] => Array
(
[11] => Eagles
)

[26] => Array
(
[12] => @ Eagles
)

[2] => Array
(
[13] => Eagles
[15] => @ Eagles
)

[13] => Array
(
[14] => @ Eagles
)

)

I have a MySQL table with a list of albums and there is a field called "views" with the number of views each album has received. I'm looking to generate an array of all the albums in the table and sort the array by the number of views (descending). I have a list of functions defined in a ContentController.php file. I created a new function called build_albumlist, which I've pasted below. The function "get_ip_log" already exists and works, and I used it as a template to create the "build_albumlist" function:





   public function build_albumlist(){
      return $this->select_raw("SELECT * FROM albums WHERE deleted = '0' ORDER BY views DESC",array(),'all');
   }

public function get_ip_log(){
      return $this->select_raw("SELECT * FROM sessions ORDER BY ID DESC",array(),'all');
   }


When I use the function, I get this warning:


Warning: mysql_real_escape_string() expects parameter 1 to be string, array given inC:\xampp\htdocs\Controllers\DBController.php on line 10

[font=cabin, 'trebuchet ms', helvetica, arial, sans-serif]The "select_raw" function that I used in "build_albumlist" is defined in the DBController.php file, and is defined as below:[/font]
private function clean_array($params){
      $final=array();
      foreach($params as $param){
         $final[]=mysql_real_escape_string($param);
      }
      return $final;
   }
   public function select_raw($query,$params,$type=''){
      $query=str_replace("?","'%s'",$query);
      $final_query= call_user_func_array('sprintf', array_merge((array)$query, $this->clean_array($params)));
      if($type==''){
         $result=mysql_query($final_query) or die(mysql_error());
         return mysql_fetch_assoc($result);
      }
      elseif($type=='all'){
         $result=mysql_query($final_query) or die(mysql_error());
         $final=array();
         while($row=mysql_fetch_assoc($result)){
            $final[]=$row;
         }
         return $final;
      }


Does anyone know why the "build_albumlist" function is generating this warning, while the "get_ip_log" is not? Any help would be great, as I am obviously pretty new to this.

Hi all,

I have the following MySQL insert query:
Code: [Select]
$insert= mysql_query   ("INSERT INTO tablename (column1,`".$EXPfields."`) VALUES ('$something','".$EXPvalues."')");
where $EXPfields is an array of table-field-names and $EXPvalues is an array of table-field-values.

Now I want to write an equivalent query, but using UPDATE instead of INSERT INTO, but I don't want to write out all the field names/values separately, but again want to use $EXPfields and $EXPvalues.

So something like this:
Code: [Select]
$update = mysql_query ("UPDATE tablename
                     SET (column1,`".$EXPfields."`) = ('$something','".$EXPvalues."')
                     WHERE ....
                  ");

Is this possible? If so, what is the proper syntax? Thanks!

Hi All,

First time posting here. I've googled the problem, but can't seem to find a response that's the same.

All I want to do is have a list of id numbers and for each id number in the array, submit a MySQL query to retrieve information relating to the id number.

When I execute the code below however, I end up with only the last item in the array being printed in the echo statement.

Any clues?

Thanks,

Code: [Select]
// get array of ids
$ids = getIDs($ids);

// loop through input list
foreach ($ids as &$id) {
getVarDetails($id);
}

function getVarDetails($local)
{
$con = mysql_connect('localhost:3306', 'root', '********');
if (!$con)
{
die('Could not connect: ' . mysql_error());
}

// set database as Ensembl
mysql_select_db("Ensembl", $con);

$result = mysql_query("SELECT * FROM variations WHERE name = '$local' LIMIT 1");
$row = mysql_fetch_array($result)

while($row = mysql_fetch_array($result))
{
echo $row['name'] . " " . $row['id'];
echo "<br />";
}

// close connection
mysql_close($con);
}

If you also have any feedback on my code, please do tell me. I wish to improve my coding base.

Basically when you fill out the register form, it will check for data, then execute the insert query. But for some reason, the query will NOT insert into the database. In the following code below, I left out the field ID. Doesn't work with it anyways, and I'm not sure it makes a difference.

Code:

  Code: [Select]
mysql_query("INSERT INTO servers (username, password, name, type, description, ip, votes, beta) VALUES ($username, $password, $name, $server_type, $description, $ip, 0, 1)");
Full code:

Code: [Select]
<?php
include_once("includes/config.php");
 
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<title><? $title; ?></title>
<meta http-equiv="Content-Language" content="English" />
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8" />
<link rel="stylesheet" type="text/css" href="style.css" media="screen" />
</head>
<body>
 
<div id="wrap">
 
<div id="header">
<h1><? $title; ?></h1>
<h2><? $description; ?></h2>
</div>
 
<? include_once("includes/navigation.php"); ?>
 
<div id="content">
<div id="right">
<h2>Create</h2>
<div id="artlicles">
<?php
 
if(!$_SESSION['user'])
{
 
        $username = mysql_real_escape_string($_POST['username']);
        $password = mysql_real_escape_string($_POST['password']);
        $name = mysql_real_escape_string($_POST['name']);
        $server_type = mysql_real_escape_string($_POST['type']);
        $description = mysql_real_escape_string($_POST['description']);
 
        if(!$username || !$password || !$server_type || !$description || !$name)
        {

        echo "Note: Descriptions allow HTML. Any abuse of this will result in an IP and account ban. No warnings!<br/>All forms are required to be filled out.<br><form action='create.php' method='POST'><table><tr><td>Username</td><td><input type='text' name='username'></td></tr><tr><td>Password</td><td><input type='password' name='password'></td></tr>";
        echo "<tr><td>Sever Name</td><td><input type='text' name='name' maxlength='35'></td></tr><tr><td>Type of Server</td><td><select name='type'>
       
        <option value='Any'>Any</option>
        <option value='PvP'>PvP</option>
        <option value='Creative'>Creative</option>
        <option value='Survival'>Survival</option>
        <option value='Roleplay'>RolePlay</option>
       
        </select></td></tr>
       
        <tr><td>Description</td><td><textarea maxlength='1500' rows='18' cols='40' name='description'></textarea></td></tr>";
        echo "<tr><td>Submit</td><td><input type='submit'></td></tr></table></form>";
        }
        elseif(strlen($password) < 8)
        {
        echo "Password needs to be higher than 8 characters!";
        }
        elseif(strlen($username) > 13)
        {
                echo "Username can't be greater than 13 characters!";
        }
        else
        {
                $check1 = mysql_query("SELECT username,name FROM servers WHERE username = '$username' OR name = '$name' LIMIT 1");
               
                if(mysql_num_rows($check1) < 0)
                {
                        echo "Sorry, there is already an account with this username and/or server name!";
                }
                else
                {
                        $ip = $_SERVER['REMOTE_ADDR'];

                        mysql_query("INSERT INTO servers (username, password, name, type, description, ip, votes, beta) VALUES ($username, $password, $name, $server_type, $description, $ip, 0, 1)");
                        echo "Server has been succesfully created!";
                }
        }
       
}
else
{
        echo "You are currently logged in!";
}
 
?>
</div>
</div>
 
<div style="clear: both;"> </div>
</div>
 
<div id="footer">
<a href="http://www.templatesold.com/" target="_blank">Website Templates</a> by <a href="http://www.free-css-templates.com/" target="_blank">Free CSS Templates</a> - Site Copyright MCTop
</div>
</div>
 
</body>
</html>

I'm restarting this under a new subject b/c I learned some things after I initially posted and the subject heading is no longer accurate.

What would cause this behavior - when I populate session vars from a MYSQL query, they stick, if I populate them from an MSSQL query, they drop.

It doesn't matter if I get to the next page using a header redirect or a form submit.  I have two session vars I'm loading from a MYSQL query and they remain, the two loaded from MSSQL disappear. 

I have confirmed that all four session vars are loading ok initially and I can echo them out to the page, but when the application moves to next page via redirect or form submit, the two vars loaded from MSSQL are empty. 

Any ideas?

A friend of mine must of changed something on the site while I was asleep last night and now all the site says when you go to it is:

Error
Database query error

Warning: mail() [function.mail]: SMTP server response: 530 SMTP authentication is required. in C:\xampp\htdocs\inc\utils.inc.php on line 449


I'm not exactly sure what he did since I can't contact him. Can anyone help me fix this?

I have been pulling my hair out for the lasy 3 hours i am trying to update a MySql table but i cant get it too work, i just keep getting MySql error #1064 - You have an error in your SQL syntax; if i just update 1 field it works fine but if i try to update more than 1 field it dosent work, Help Please!


<?php
$root = $_SERVER['DOCUMENT_ROOT'];
require("$root/include/mysqldb.php");
require("$root/include/incpost.php");
$con = mysql_connect("$dbhost","$dbuser","$dbpass");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("$dbame", $con);


mysql_query("UPDATE Reg_Profile_p SET
build='$build'
col='$col'
size='$size'
WHERE uin = '$uinco'");
?>

How can i save array from inputs, witch is $igraci and it have 5 values, to mysql base, i tryed but in mysql it shows Array.
This is form:
        <form action="dodaj_klan.php" method="post" >
            <p>
            <label>Naziv klana:</label>
            <input name="naziv" type="text" size="20%" />
            <label>Web Sajt:</label>
            <input name="website" type="text" size="20%" />
            <label>E-Mail:</label>
            <input name="email" type="text" size="20%" />
            <br /><br />
            <label>Igraci klana(5):</label>
            <input name="lider" type="radio" value="1" />
            <input name="igraci[]" type="text" size="20%" /><br />
            <input name="lider" type="radio" value="2" />
            <input name="igraci[]" type="text" size="20%" /><br />
            <input name="lider" type="radio" value="3" />
            <input name="igraci[]" type="text" size="20%" /><br />
            <input name="lider" type="radio" value="4" />
            <input name="igraci[]" type="text" size="20%" /><br />
            <input name="lider" type="radio" value="5" />
            <input name="igraci[]" type="text" size="20%" />
            <label><i>(obelezi lidera klana)</i></label>
            <br /><br />
            <input class="button" type="submit" name="submit" value="Dodaj" />&nbsp;&nbsp;
            </p>
        </form>
Now i wonna save this igraci[] array in mysql, i tried like this but it doesn't work:
$igraci = $_POST['igraci'];
$query = "INSERT INTO klanovi (naziv, website, email, igraci) VALUES ('{$naziv}', '{$website}', '{$email}', '{$igraci}')";
$result = mysql_query($query, $connection);

How can i do this?
Thanks..

My script IS working, but I can't get around a blank array error when no errors exist.

Below is my code, and as you can see, I am being handed this:
Notice: Undefined variable: error in C:\wamp\www\php\form_validation.php on line 19 as a result of my ... if (is_array($error)) { ... .. I could do if (@is_array($error)) { (note the @), but I hate using that thing... I've tried several things with no luck, so any ideas welcome at this point.


<?php
  if (isset($_POST['set_test'])) {
      if (!preg_match("/^[A-Za-z' -]{1,50}$/", $_POST['first_name'])) {
          $error[] = "Please enter a valid First Name";
      }
      if (!preg_match("/^[A-Za-z' -]{1,50}$/", $_POST['last_name'])) {
          $error[] = "Please enter a valid Last Name";
      }
      if (is_array($error)) {
          foreach ($error as $err_message) {
              echo $err_message . "<br />";
          }
      }
  }
?>