I am getting tired trying to see what is wrong. I have two php. From the first I am sending a variable 'select1' (basically the id) to the second and than I want to update that record uploading a pdf file.
$id = "-1";
if (isset($_GET['select1'])) {
$id = mysql_real_escape_string($_GET['select1']);
}
if(isset($_POST['Submit'])) {
$my_upload->the_temp_file = $_FILES['upload']['tmp_name'];
$my_upload->the_file = $_FILES['upload']['name'];
$my_upload->http_error = $_FILES['upload']['error'];
if ($my_upload->upload()) { // new name is an additional filename information, use this to rename the uploaded file
mysql_query(sprintf("UPDATE sarcini1 SET file_name = '%s' WHERE id_sarcina = '%s'", $my_upload->file_copy, $id));
}
}
If I put a line with a valid id, like:
$id = 14;
it is working. What I am doing wrong? Thank you!
If you need to accept both post & get, then you should try something like the code below to retrieve the variable.
$var = 'select1';
if( isset( $_POST[$var] ) ) {
$id = $_POST[$var];
} else if( isset( $_GET[$var] ) ) {
$id = $_GET[$var];
} else {
$id = -1;
}
You are using both GET and POST at the same time. As far as I can see, this condition is not returning True
if (isset($_GET['select1']))
Edit: If you don't find any answer in above; maybe some more information/code can help getting to a solution.
Related
I need to stop inserting in the WordPress database a value if already exists. What I tried and I don't know why it's not working. I'm using the next code in functions.php and I have and ajax call. Can someone make some light on this please.
My code
$i_can_add_value = true;
$error_reason = false;
$published_values = $wpdb->get_results(
"SELECT ID FROM ".$wpdb->posts." WHERE post_author = %d AND post_status IN ('draft','publish') AND post_type = 'custom' ", $current_user->ID
);
if (!empty($published_values)){
foreach ($published_values as $id_already_published){
$temp_custom = get_post_meta($id_already_published->ID, 'custom', true);
if ($temp_custom === $description && $description !='') {
$i_can_add_value = false;
$error_reason = 'You already added that!';
}
}
}
if($i_can_add_value){
wp_insert_post($my_post);
} else {
echo "Error: ".$error_reason;
}
First of all you need to be sure that you code is working at all when ajax is called.
If this code is working then you need to check three conditions in your code.
1) if (!empty($published_values)
$published_values can be empty because of the incorrect sql query.
2) $temp_custom
Can also be empty. Check it.
3) $description
And this variable can be empty as well.
I have some problem passing array of data in Php using session:
In my class, I have this function:
public function check_dupID($id)
{
$this->stmt="select id from student where id='$id'";
$this->res = mysql_query($this->stmt);
$this->num_rows = mysql_num_rows($this->res);
if($this->num_rows == 1)
{
while($this->row = mysql_fetch_array($this->res))
{
$dup_id = $this->row[0];
return $dup_id;
}
}
}
This code checks for duplicate ID of a student. Now, the reason for this is that I uploaded the file in the database using ".csv" file.
Before Uploading it I put all the data in the csv file into an array. Now, I check it one by one:
//created the variable for checking
$id = $check->check_dupID($array[0]);
if($id == TRUE)
{
session_start();
$_SESSION['id']; = $id;//passing it to the next back to array again.
$redirect->page($error);
}
Now when it redirect it gives me the result of an "Array" and if I parse it using foreach it shows only one id but there is more than of that. I want to know how to display it all.
This code will return the duplicate id when it is found for the first time. If you want to return all the duplicate ids, then add them into an array using the while loop and then return it.
To do this modify your code like this:
public function check_dupID($id)
{
$this->stmt="select id from student where id='$id'";
$this->res = mysql_query($this->stmt);
$this->num_rows = mysql_num_rows($this->res);
$dup_id = array();// added here to prevent invalid argument supplied to foreach loop error if nothing is found
if($this->num_rows >1)
{
while($this->row = mysql_fetch_array($this->res))
{
$dup_id[] = $this->row[0];// adding duplicates into the array
}
return $dup_id;
}
}
This code fixes my problem:
if($id == TRUE)
{
$get_id = array();
$get_id[] = $id;
session_start();
$_SESSION['id']; = $get_id;//passing it to the next back to array again.
$redirect->page($error);
}
When I access the wrong function in my controller, a 404 error page is working well. But, when i access url like 'http://example.com/model/detail/116', which 116 is wrong number [are not in the database], my 404 error page not working.
I have this code in my controller:
public function detail()
{
$id['id_galeri'] = $this->uri->segment(3);
$detail = $this->app_model->getSelectedData("tbl_galeri",$id);
foreach($detail->result() as $d)
{
$bc['jdl'] = "View";
$bc['id_galeri'] = $d->id_galeri;
$bc['nama'] = $d->nama;
$bc['foto'] = $d->foto;
$bc['deskripsi'] = $d->deskripsi;
$bc['stts_input'] = "deskripsi";
}
if($this->uri->segment(3) == '' && $id['id_galeri'] == FALSE)
{
$segment_url = 0;
}else{
if(!is_numeric($this->uri->segment(3)) || !is_string($this->uri->segment(3))){
redirect('404');
}else{
$segment_url = $this->uri->segment(3);
}
}
$this->load->view('frontend/global/bg_top');
$this->load->view('frontend/page/bg_view_model',$bc);
$this->load->view('frontend/global/bg_footer');
}
Sorry for my bad english, please help :-)
Thank you..
Instead of:
redirect('404');
try using CodeIgniter's native:
show_404('page');
EDIT
Try this code, a bit cleaned up and the checks are done before they're saved for views use.
public function detail()
{
$id['id_galeri'] = $this->uri->segment(3);
// Check if the supplied ID is numeric in the first place
if ( ! is_numeric($id['id_galeri']))
{
show_404($this->uri->uri_string());
}
// Get the data
$detail = $this->app_model->getSelectedData("tbl_galeri",$id);
// Check if any records returned
if (count($detail->result()) === 0)
{
show_404($this->uri->uri_string());
}
foreach($detail->result() as $d)
{
$bc['jdl'] = "View";
$bc['id_galeri'] = $d->id_galeri;
$bc['nama'] = $d->nama;
$bc['foto'] = $d->foto;
$bc['deskripsi'] = $d->deskripsi;
$bc['stts_input'] = "deskripsi";
}
/**
* Here do whatever you want with the $segment_url which doesn't seem to be
* used in your code
*/
$this->load->view('frontend/global/bg_top');
$this->load->view('frontend/page/bg_view_model',$bc);
$this->load->view('frontend/global/bg_footer');
}
I'm writing an application that allows users to upload reference letters for potential employees.
Every reference is sent an email containing a unique string at the end of the url. So, for example, an address would look similar to: www.mywebaddress?url=503241a20b5085_18720621.
To determine if the unique string is valid (i.e. exists in the database) I need to do a query search. However, when a reference attempts to access the URL he needs to answer a security question. So, I also need to check if the answer is valid, if he has previously uploaded, etc to determine what page to redirect him to.
But because of the query, my code requires the user to click "Submit" twice. This is really annoying, but I'm not sure how to fix it.
Here is a relevant excerpt of my code:
if ( isset ($_GET['url']) ) {
$query = "SELECT * FROM ref_info WHERE url='" . $_GET['url'] . "'";
$result = $db->execute($query);
if ( empty ($result) ) {
//error message
} else {
$url = $_GET['url'];
if ( $_SESSION['validated'] ) {
if ( $result[0]['uploaded'] ==1 ) {
$_SESSION['uploaded'] =true;
} else {
$_SESSION['uploaded'] =false;
}
include_once("process_upload.php");
} else {
if ( empty($result[0]['answer']) ) {
include_once("security.php");
} else {
include_once("security_check.php");
}
}
}
}
Is there anything I can do so that the form only needs to be submitted once?
Thanks in advance for any suggestions!!
if ( isset ($_GET['url']) ) {
$query = "SELECT * FROM ref_info WHERE url='" . $_GET['url'] . "'"; //that is really not secure. Take a look at mysql_real_escape_string or something like that in yor $db
$result = $db->execute($query);
if ( empty ($result) ) {
//error message
} else {
$url = $_GET['url'];
if (empty($result[0]['answer']) ) { // page is just opened,form is not yet submitted - ask sequrity question
include_once("security.php");
} else { //oh. Submit is done - let me check if it is Ok
include_once("security_check.php");
}
if ( $_SESSION['validated'] ) { //validation is Ok? Yeah. Answer is not posted yet, so validation fails and we do nothing. Just show a form with a question
if ( $result[0]['uploaded'] ==1 ) {
$_SESSION['uploaded'] =true;
} else {
$_SESSION['uploaded'] =false;
}
include_once("process_upload.php");
}
}
}
I'm trying to write a simple function to construct field names for a form. It works fine if at least one value is selected in a multi-select list but if nothing is selected I get an Undefined index error. Here is what I have:
function mcFieldName($mcFieldName){
$mcField = $_POST[$mcFieldName];
if( !is_array($mcField) ){
if( !empty($mcField) ){
return $mcField;
}else{
return 'n/a';
}
}
if( is_array($mcField) ){
$mcFieldArray = implode(',', $mcField);
return $mcFieldArray;
}
}
$MultiSelect = mcFieldName('mcMultiSelect');
// test
echo $MultiSelect . '<br/>';
Thank you!
You just need to protect yourself from reading a key that does not exist in $_POST:
$mcField = isset($_POST[$mcFieldName]) ? $_POST[$mcFieldName] : null;
Before you try to access an array item make sure it exists with using isset():
if (isset($_POST[$mcFieldName])) {
$mcField = $_POST[$mcFieldName];
...
}