How to check all values present in the column 'pincode'? - php

This code only reads the first value present in the column. If the value posted in the html form matches the first value, it inserts into the database. But I want to check all the values in the column and then take the respective actions.
For example, if i give input for 'ppincode' and 'dpincode' as 400001, it accepts. but if i gave 400002, 400003,..... it displays the alert even if those value are present in the database
DATABASE:
pincode <== column_name
400001 <== value
400002
400003
400004
...
also i tried this
$query = "SELECT * FROM pincodes";
$result = mysqli_query($db, $query);
$pincodearray = array();
if (mysqli_num_rows($result) > 0) {
while($row = mysqli_fetch_assoc($result)){
$pincodearray[] = $row;
}
}

If I understand well - you want to compare value from POST request with all retrieved records saved in DB and if it matches - perform action.
If so, I would recommend using for(each) loop. Example:
if( !empty($row){
foreach( $row as $key ){
if($key['pincode'] == $ppincode && $key['pincode'] == $dpincode){
// your action goes here
}
}
}
Additional tip: use prepared statements :)

SELECT count(*) FROM table WHERE ppincode=ppincode AND bpincode=bpincode
if this return 0 then insert or else show alert.

Related

Remove Array From Json in PHP

hi i have a backend with php in cpanel and i have a problem with one of jsons . this is part of my php code :
...
}elseif ($work == "dollardate") {
$query3 = "SELECT * FROM tabl_dollar_date";
$result3 = $connect->prepare($query3);
$result3->execute();
$out3 = array();
while ($row3 = $result3->fetch(PDO::FETCH_ASSOC)) {
$record3 = array();
$record3["dollar"] = $row3["dollar"];
$record3["date"] = $row3["date"];
array_push($out3, $record3);
}
echo json_encode($out3);
}
?>
this code show this in json :
[
{
"dollar":"15000",
"date":"1397-12-12"
}
]
how can remove array from json and show the json like this :
{
"dollar":"15000",
"date":"1397-12-12"
}
Easiest way (according his code):
change line
echo json_encode($out3);
to
echo json_encode($out3[0]);
One solution is that if you just want the latest value (in case there are multiple records in the table), then change the SELECT to order by date descending also set LIMIT to 1 to only get the 1 record anyway, and remove the loop to fetch the data and just fetch the 1 record...
$query3 = "SELECT `date`, `dollar`
FROM `tabl_dollar_date`
ORDER BY `date` desc
LIMIT 1";
$result3 = $connect->prepare($query3);
$result3->execute();
$row3 = $result3->fetch(PDO::FETCH_ASSOC);
echo json_encode($row3);
As you know which fields you want from the SELECT, it's good to just fetch those fields rather than always using *. This also means that as the result set only contains the fields your after, you can directly json_encode() the result set rather than extracting the fields from one array to another.

How to block inserting empty strings in MySQL

I have this update statement (PHP code):
$sql1="UPDATE `utilizatori` " .
"SET utilizator='$utilizator', parola='$parola1', nume='$nume', " .
"`prenume='$prenume', varsta='$varsta', localitate='$localitate'` ";
WHERE parola='".$_SESSION['parola']."'";
This will update some MySQL table fields via an html form. The user wants to change just his name for instance. He completes just name field, then he presses submit. The data is sent into the table with the UPDATE statement above.
The problem is that it also updates the table with blank values that user didn't complete. I don't want the blank values to be added.
How can I block the blank values to be sent into the table?
If you really wanted to do this in the update, you can change the set statement to something like:
set utilizator = (case when '$utilizator' <> '' then '$utilizator' else utilizator end),
. . .
This will use the previous value if the new one is blank.
You can also do this at the application level by just updating the fields that have changed.
And, you should use parameterized queries rather than directly substituting values into a string. That is another issue, though.
You can do two things to solve this issue. One is to preload the data in the form. So when the user change his name, the other fields are already loaded with the original information.
The second option is to create an update query based on the fields have a value.
Example of option 1:
<?php
//
//GET THE DATA FROM A SELECT QUERY HERE
//FOR EXAMPLE: $sql = "SELECT * FROM `utilizatori` WHERE parola='".$_SESSION['parola']."'";
//Put the data of the sql row in a variable e.g. $sqlRow.
?>
<!--Use variable in your form!-->
<form>
...
...
<input name="nume" value="<?=$sqlRow['nume']?>"/>
<input name="utilizator" value="<?=$sqlRow['utilizator']?>"/>
...
...
</form>
Example of option 2:
<?php
//Catch post data
if($_POST)
{
$updateString = "";
foreach($_POST as $inputField => $inputValue)
{
if($inputValue != "")
{
$updateString .= $inputField." = '".$utilizator."',";
}
}
//Strip last ,
$updateString = substr($updateString,0,-1);
if($updateString != "")
{
//Your query would be
$sql1 = "UPDATE `utilizatori` SET ".$updateString." WHERE parola='".$_SESSION['parola']."'";
}
}
?>
$updateClauseArr = Array();
foreach($_REQUEST as $key => $val){
if(is_numeric($val)){
$updateClauseArr[] = '$key = '.(int) $val;
}else{
$updateClauseArr[] = "$key = '".htmlentities($val,ENT_QUOTES,'UTF-8')."'";
}
}
if(sizeof($updateClauseArr) > 0){
$updateSet = implode(',' ,$updateClauseArr);
$sql1="UPDATE `utilizatori` SET ".$updateSet." WHERE parola='".$_SESSION['parola']."'";
}
See what field values have been submitted by the user. then iterate in a loop for the fields that have value to make variable to be concatenated to the update query.

how to identify unsaved value from database using php mysql

I have to identify unsaved value from mysql table using php and mysql, for example i using table named as numtab and i have already stored some numbers 1, 3, 4, 7, 23, 12, 45 in numb column.
now i have generated one new number randomly(for example 23) and i have to check this number with already stored numbers,
if 23 is exist in the table mean i have to generate another one new number and have to check once again with stored values, this process have to continue till finding unsaved number.
if generated value is not exist in table mean can stop the process and can store this number in table.
here below the format i am currently using
$numb=23;
$qryb="select * from numtab where numb='$numb'";
$results=mysql_query($qryb)or die("ERROR!!");
if(mysql_num_rows($results) == 1)
{
$numb=rand(1,100);
mysql_query("insert query");
}
the problem is above the code is validation once only, its not verifying second time. i think if using for or while loop mean can solve this problem, but i dont know how to do looping, so help me to solve this problem.
You can use in clause like this :
$qryb="select * from numtab where numb in('$numb')";
$results=mysql_query($qryb)or die("ERROR!!");
$count = mysql_num_rows($results);
if ($count > 0) {
echo "number exist in db";
} else {
echo "number does not exist in db";
}
You could make a while() loop to check if the numbers exist in your database. You could also retrieve all numbers from the database, store them in an array and check if the generated number exists within that array.
The first option would be something like this:
do {
$numb=rand(1,100);
$qryb="select * from numtab where numb='$numb'";
$results = mysql_query($qryb) or die("ERROR!!");
} while(mysql_num_rows($results) >= 1)
mysql_query("insert query");
The second option would be something like this:
$query = mysql_query("SELECT DISTINCT(numb) as numb FROM numtab");
// set array
$array = array();
// look through query
while($row = mysql_fetch_assoc($query)){
// add each row returned into an array
$array[] = $row['numb'];
}
do
{
$numb = rand(1,100);
}
while(!in_array ( $numb , $array) ;
mysql_query("insert query");

Trying to get a page to populate data from a database

I'm trying to write a script that gets data from a sql server based on the id of the entry in my data base. when I try to access the page using the link with the id of the entry it returns as if it does not recognize the id. Below is the php code :
<?php
require('includes/config.inc.php');
require_once(MYSQL);
$aid = FALSE;
if (isset($_GET['aid']) && filter_var($_GET['aid'], FILTER_VALIDATE_INT, array('min_range' => 1)) ) {
$aid = $_GET['aid'];
$q = "SELECT aircraft_id, aircraft_name AS name, aircraft_type AS type, tail_number AS tn FROM aircraft USING(aircraft_id) WHERE aircraft_id = $aid";
$r = mysqli_query($dbc, $q);
if (!(mysqli_num_rows($r) > 0)) {
$aid = FALSE;
}
}// end isset tid
if ($aid) {
while ($acdata = mysqli_fetch_array($r, MYSQLI_ASSOC)){
echo'<h2>'. $acdata['tail_number'] .'</h2>';
}
} else {
echo '<p>This pages was accessed in error.</p>';
}
?>
Any hints?
Try to var_dump($q); before $r = mysqli_query($dbc, $q); to inspect your query and then just execute it through phphmyadmin or in MySQL server terminal directly and see what does it return.
Update:
Use var_dump($q);die(); to stop script from executing after dumping.
You are using a field alias in your query, so you must use that in your echo to:
echo'<h2>'. $acdata['tn'] .'</h2>';
Get rid of the USING(aircraft_id), it causes an error and your query doesn't execute.
"SELECT aircraft_id, aircraft_name AS name, aircraft_type AS type, tail_number AS tn FROM aircraft WHERE aircraft_id = $aid"
I guess it's a leftover from a previous version of the query? Using (id) is a shortcut for
FROM
foo
INNER JOIN bar ON foo.id = bar.id
It can be used when the tables to be joined are joined on columns which have the same name. Just shorter to write.
Since you are not joining you have to remove it.

PHP - SQL select first column with no data

How do I find the first column with no data?
//column pet1='dog'
//column pet2=''
//column pet3=''
//column pet4='cat'
if($F=="getempty"){
$uid=$_GET["uid"];
$sql = mysql_query("SELECT DISTINCT pet1,pet2,pet3,pet4 FROM a WHERE uid='".$uid."' AND pet1='' OR pet2='' OR pet3=''OR pet4=''");
while($column=mysql_fetch_field($sql)){
$empty=$column->name;
echo json_encode(array("empty"=>$empty));
}
}else{}
I keep trying but so far it just returns all column names if one is empty
pet1,pet2,pet3,pet4
DISTINCT will combine the rows which are exactly alike (having all the selected columns the same) and return them.
To get the name of the first empty column, use a query like:
SELECT
CASE
WHEN pet1 = '' THEN 'pet1'
WHEN pet2 = '' THEN 'pet2'
WHEN pet3 = '' THEN 'pet3'
WHEN pet4 = '' THEN 'pet4'
ELSE NULL
END AS firstempty
FROM a
WHERE uid = '$uid';
The idea is that the CASE statement checks each one in order, and finding an empty one returns the column name without checking the rest of them.
Don't forget to escape $uid with mysql_real_escape_string() before passing it to the query.
$uid = mysql_real_escape_string($uid);
Since we're now getting the column name inside a field called firstempty, we'll also need to change the way it's fetched:
while($row = mysql_fetch_assoc($sql)){
$empty = $row['firstempty'];
echo json_encode(array("empty"=>$empty));
}

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