The following code shows an error
"Undefined offset: 2" on lines 22, 28, and 29
$sql = "SELECT email FROM CommercialEmails WHERE dripid = 1 AND sent='a'";
if ($resultsd1 = mysqli_query($conn, $sql )) {
$affectedrows = mysqli_num_rows($resultsd1);
while ($row = mysqli_fetch_row($resultsd1)){
$results = $row[0];
global $results;
}
}
$broken = explode(' ', $results);
$hi = 0;
$hello = 0;
a:
/**Line 22 **/ if (substr($broken[$hi], -4) == "com," && $broken[$hi] == "qwert"){
$hey[$hi] = $broken[$hi];
$hello++;
}
If(substr($broken[$hi], -4) !== "com,"){ // line 28
$hey[$hi] = $broken[$hi]; //Line 29
}
$hi++;
if ($hi == $affectedrows){
if (!isset($hey)){
echo "There are no emails";
} else {
foreach( $hey as $key => $value){
echo $value;
}
echo $hey;
}
}else{
goto a;
}
I am not sure what you are trying to achieve, but to get your code just working try this way:
$sql = "SELECT email FROM CommercialEmails WHERE dripid = 1 AND sent='a'";
$results = array();
$hey = array();
if ($resultsd1 = mysqli_query($conn, $sql )) {
$affectedrows = mysqli_num_rows($resultsd1);
while ($row = mysqli_fetch_row($resultsd1)){
$results[] = $row;
If(substr($row[0], -4) !== "com,"){
$hey[] = $row[0];
}
}
}
if (count($hey)==0){
echo "There are no emails";
} else {
foreach( $hey as $value){
echo $value;
}
echo $hey;
}
As you can see I completely removed many of your variables you don't need them until you can explain your goals.
And I've removed your weird condition if:
if (substr($broken[$hi], -4) == "com," && $broken[$hi] == "qwert"){
$hey[$hi] = $broken[$hi];
$hello++;
}
Because there is no such value of $broken[$hi] that can be equal 'qwert' and to have 'com' as substring inside. So this codition is always false and we can delete it.
Try to explain your goals. Hope I can help you.
It means that the $broken array does not have 3 elements (ofset index = 2).
I can not see where that $hi var is being set to 2 but it is somewhere in code you have not shown I guess.
Instead of global $results; inside the while loop, before the while loop starts just declare $results as an empty array $results = array();
Then after the while loop ends do print_r($results); to see if it has the contents you expect.
Related
There are three "BC" in the result_category but I am only getting 1 result. This is for a personality quiz. Please Help. I also tried $query = "SELECT result FROM quiz_map where result_category = 'BC'"; but still, only 1 result is showing.
$result = mysqli_query($link, $query);
$cat_a = $cat_b = $cat_c = $cat_d = $cat_e = 0;
while($row = mysqli_fetch_array($result, MYSQLI_BOTH)) {
$cat = $row['category'];
if ($cat == "A") {
$cat_a += 1;
} elseif ($cat == "B") {
$cat_b += 1;
} elseif ($cat == "C") {
$cat_c += 1;
} elseif ($cat == "D") {
$cat_d += 1;
} elseif ($cat == "E") {
$cat_e += 1;
}
}
$array = array('A' => $cat_a, 'B' => $cat_b, 'C' => $cat_c, 'D' => $cat_d, 'E' => $cat_e);
$str = '';
foreach ($array as $i => $value) {
if ($value >= 6) {
$str = $i;
break;
} elseif ($value >= 2) {
$str .= $i;
}
}
$var = sort($array);
$query = "SELECT result FROM quiz_map where result_category = '$str' LIMIT 1";
$result = mysqli_query($link, $query);
$row = mysqli_fetch_array($result);
echo $row[0];
?>
There's many things wrong with your code!
As pointed by #IsThisJavascript and #Cashbee:
You are executing a query with a LIMIT 1 statement, it will only return one record.
As pointed by myself:
Doing echo $row[0] will have the same result, if you are only echoing the first value of the array you can't expect to have multiples can you?
As pointed by #IsThisJavascript:
You need to loop the results array, like so:
while($row = mysqli_fetch_array($result)){
echo $row['result_category'];
}
Consider switching the query from a '=' to a '%like%' statement, to maximize results if you want to get the values that partionaly cointain the string.
I'm want string out of the column data.
But it failed.
<?php
$conn = mysql_connect("localhost", "nantawat", "12345678") or die(mysql_error());
$select_db = mysql_select_db("my_db", $conn) or die(mysql_error());
$select_tbl = mysql_query("SELECT * FROM log", $conn);
while ($fetch = mysql_fetch_object($select_tbl)) {
$r = $fetch->data;
$i = explode(",", $r);
if (!isset($i[1])) {
for ($j = 0; $j <= 200; $j++) {
$i[$j] = null;
}
}
$name = $i[0];
$mama = $i[1];
$no = $i[2];
$a = $i[3];
$b = $i[4];
echo $name . "</br>";
echo $mama . $no . $a . $b . "</br>";
}
while ($data = mysql_fetch_object($select_tbl)) {
echo $data->data . "<br>";
}
?>
But i want output =
bus
car
bike
aabus
car
bike
aabus
car
bike
aabus
ddd
ee
And i not
Notice: Undefined offset: 3 in C:\xampp\htdocs\logs\New folder
(2)\explode.php on line 21
Notice: Undefined offset: 4 in C:\xampp\htdocs\logs\New folder
(2)\explode.php on line 22
Thank You.
You should just do what you want to do.
You want to connect to database then do it:
$conn = mysql_connect("localhost", "nantawat", "12345678") or die(mysql_error());
I suggest you to use mysqli library instead of mysql (mysql is deprecated in new php versions and totally removed in php7)
$conn = mysqli_connect("localhost", "nantawat", "12345678", "my_db") or die(mysql_error());
You want to query on log table, then do it:
$select_tbl = mysqli_query($conn, "SELECT * FROM log");
You want to fetch info from your result, then do it:
while ($row = mysqli_fetch_array($select_tbl)) {
echo $row['id_user'];
echo $row['id_doc'];
echo $row['date'];
echo $row['data'];
}
You want to explode data, then do it:
while ($row = mysqli_fetch_array($select_tbl)) {
echo $row['id_user'];
echo $row['id_doc'];
echo $row['date'];
$data = explode(',', $row['data']);
foreach ($data as $d) {
if ($d !== '') { // because before first comma or after last can be empty
echo $d . PHP_EOL;
}
}
}
If you want to save database result in variables:
If you are getting only one row of database, you can save them in variables directly:
$id_user = '';
$id_doc = '';
$date = '';
$data = array();
id ($row = mysqli_fetch_array($select_tbl)) {
$id_user = $row['id_user'];
$id_doc = $row['id_doc'];
$date = $row['date'];
$tempData = explode(',', $row['data']);
foreach ($tempData as $d) {
if ($d !== '') {
$data[] = $d;
}
}
}
And if you have multiple rows of database you need to save them all in a total array:
$array = array();
id ($row = mysqli_fetch_array($select_tbl)) {
$id_user = $row['id_user'];
$id_doc = $row['id_doc'];
$date = $row['date'];
$data = array();
$tempData = explode(',', $row['data']);
foreach ($tempData as $d) {
if ($d !== '') {
$data[] = $d;
}
}
$array[] = array(
'id_user' => $id_user,
'id_doc' => $id_doc,
'date' => $date,
'data' => data,
);
}
And finally use this to see what structure your final array has:
echo '<pre>';
pront_r($array);
echo '</pre>';
First off it is not wise to store comma seperated values in a single cell and you are using deprecated mysql_ functions. I think your solution can be found in using a foreach instead of the isset part:
while ($fetch = mysql_fetch_object($select_tbl)) {
$r = $fetch->data;
$i = explode(",", $r);
foreach ($i as $q){
echo $q . '<br/>';
}
}
If you still want to access your variables $name, $mama, $no and $ab, you can use isset for those specifically.
while ($fetch = mysql_fetch_object($select_tbl)) {
$r = $fetch->data;
$i = explode(",", $r);
if (isset($i[0])){
$name = $i[0];
echo $name . '<br>'; //only echo if it exists
}
if (isset($i[1])){
$mama= $i[1];
echo $mama. '<br>'; //only echo if it exists
}
//try it yourself for $no and $ab
}
Try:
while ($row = mysqli_fetch_array($select_tbl)) {
extract($row);
/* Using extract method can get the array key value as variable
Below variables are available
$id_user;
$id_doc;
$date;
$data; */
}
Well, basically what this code does is grab some links from a source code of a website and send them to an mp3 player.
The big problem is on the get_link function, where i want to store the urls to an array. The section where im having problems is commented.
Sorry for posting all this code but the functions are connected to each others.
function getHost($db,$id){
if(isset($_GET['id'])){
$sql1 = "SELECT host FROM mixtape WHERE id=?";
$stm = $db->prepare($sql1);
$stm->execute(array($id));
$row1 = $stm->fetch(PDO::FETCH_ASSOC);
if($row1['host']=='host1'){
$sql2 = "SELECT link1 FROM faixa WHERE id_mixtape IN(SELECT id FROM mixtape WHERE id=?)";
$stm = $db->prepare($sql2);
$stm->execute(array($id));
$rows_affected = $stm->rowCount();
$array=array();
if (count($rows_affected) > 0) {
for($i=1; $i <= $rows_affected; $i++) {
$row2 = $stm->fetch(PDO::FETCH_ASSOC);
$url=$row2['link1'];
get_Link($db,$url,$i,$rows_affected,$array);
}
}
}
}
}
function get_Link($db,$url,$pos,$rows_affect,$array){
$find = 'url:';
$reg_exUrl = "/(http|https|ftp|ftps)\:\/\/[a-zA-Z0-9\-\.]+\.[a-zA-Z]{2,3}(\/\S*)?/";
$data = file_get_contents($url);
$data = explode("\n", $data);
for ($line = 0; $line < count($data); $line++) {
if (strpos($data[$line], $find) !== false) {
$link = preg_replace($reg_exUrl,"", $data[$line]);
$v[]=$link;
}
}
if($pos!=$rows_affect-1){
$url="mylink.com/".$link."|";
}else{
$url="mylink.com/".$link."&";
}
$array[$pos]=$url;
var_dump($array); // Here says that are 3 values in the array. True
if($pos==$rows_affect-1){
var_dump($array); // Here is only showing the last value in the array. Why?
player($db,$array);
}
}
function player($db,$array){
if(isset($_GET['id'])){
foreach($array as $i=>$item){
echo $item;
}
}
}
This piece of code:
$c=0;
for ($line = 0; $line < count($data); $line++) {
if (strpos($data[$line], $find) !== false) {
$link = preg_replace($reg_exUrl,"", $data[$line]);
$v[$c]=$link;
}
}
Should be like:
$c=0;
for ($line = 0; $line < count($data); $line++) {
if (strpos($data[$line], $find) !== false) {
$link = preg_replace($reg_exUrl,"", $data[$line]);
$v[$c]=$link;
$c = $c+1; //this is missing or $c++;
}
}
OR:
for ($line = 0; $line < count($data); $line++) {
if (strpos($data[$line], $find) !== false) {
$link = preg_replace($reg_exUrl,"", $data[$line]);
$v[]=$link; //That way works too
}
}
Miguelfsf, you must first learn about variable scope.
Your problem is, you created the $array=array() (the $array show be the $row2).
You declared a new variable, but you didn't used, the $array was just to tell you, declare your array that way
$row2=array().
After that you need to do
$row2[] = $stm->fetch(PDO::FETCH_ASSOC);
Why?
Because the assoc returns a associative array, then it will do
$array => {title => "last title} = $newData => {title => "new title"}
It will replace the value
Using the [] everytime you do this it'll create a new element.
Then
{
0 => { title => "title1"}
1 => { title => "title2"}
2 => { title => "title3"}
3 => { title => "title4"}
}
I have a code where it should check if the result equals to 8 it need to show something and if not it need to show something else and all of that happens inside of a while loop.
while ($row_fpages2 = mysql_fetch_array($result_fanpage2))
{
if ( $row_fpages2['client'] != NULL ) {
//GRAPHS
$sql = "SELECT likes, date FROM statistics_pages WHERE idnum = '".$idnum."' AND page_name = '".$row_fpages2['page_name']."' ORDER BY `statistics_pages`.`date` DESC LIMIT 8";
$result2 = mysql_query($sql) or die(mysql_error());
if ($result2) {
$data = array();
while ($row = mysql_fetch_assoc($result2)) {
$data[] = $row["likes"];
}
if ($result2 == 8) {
$c_data = count($data)-1;
$final = array();
for ($i = 0; $i < $c_data; $i++) {
$final[] = getZeroResult($data[$i], $data[$i+1]);
}
$data_string = join(",", $final);
$stats = '<img src="http://chart.apis.google.com/chart?chs=240x140&cht=ls&chd=t:0,0|'.$data_string.'&chg=20,20&chls=0.75,-1,-1|6,4,1&chm=o,FF9900,1,-1,7,-1|b,3399CC44,0,1,0"></img>';
} else {
$stats = '<img src="images/stats_un.jpg"></img>';
};
} else {
print('MySQL query failed with error: ' . mysql_error());
}
echo '...';
The problem is that the first output always showing the ( == 8) (even if it is not equals to 8) instead of the else output.
Then if i have 2 or more everything comes above the first one is correct but the first one is still showing the ( == 8).
Any help?
You do the following which is incorrect:
$result2 = mysql_query($sql) or die(mysql_error());
...
if ($result2 == 8) {
The return value of mysql_query is a resource not an int. What is that you are trying to do there ?
May be you would like to use
if(strlen($result2) == 8){
...
}
instead of
if($result2 == 8){
...
}
Hope that solves your problem.
here is my code:
$i = 0;
$list4 = array();
while($row_sent = $GLOBALS['db']->sql_fetchrow($res_sent))
{
$sql_sent2 = "SELECT * FROM ".$GLOBALS['table']['sent']." WHERE `sent_id` ='".$row_sent['sent_id']."'";
//echo $sql_size2; exit;
$res_sent2 = $GLOBALS['db']->sql_query($sql_sent2);
$num = $GLOBALS['db']->sql_numrows($res_sent2);
if($num > 0)
{
while($row_sent2 = $GLOBALS['db']->sql_fetchrow($res_sent2))
{
$list4[$i]['sent_id'] = $row_sent2['sent_id']; //this line shows error
$list4[$i]['sent_name'] = $row_sent2['sent_name'];//this line shows error
$list4[$i]['sent_qty'] = $row_sent['sent_qty'];//this line shows error
}
}
else
{
$list4=0;
}
$i++;
}
try this.
declare $list4 as an array before your loop start.
$list4[] = array();
this may be useful.
Thanks.