table data(id, name)
function getData() {
$data = array();
$sql = 'Select * From data';
$query = mysql_query($sql);
if(!$query) {
echo "Error: " . mysql_error();
exit;
}
while($row = mysql_fetch_array($query)) {
$data[] = $row;
}
return $data;
}
$data = $this->getData();
foreach($data as $dt) {
echo $dt->name;
}
I get an error when I echo $dt->name;, the output is null, How do I fix it ?
$dt is not an object but a array. $dt->name should be $dt['name'].
Try:
var_dump($data);
//if its not a class then simply do
$data = getData();
Related
I am making a CRUD system for blog publications, but it's kinda strange, another developer (with more experience) looked to my coded and for him it's all right too, but this error (Notice: Trying to get property of non-object in C:\xampp\htdocs\genial\painel\inc\database.php on line 32) remains appearing.
My database code:
<?php
mysqli_report(MYSQLI_REPORT_STRICT);
function open_database() {
try {
$conn = new mysqli(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME);
return $conn;
} catch (Exception $e) {
echo $e->getMessage();
return null;
}
}
function close_database($conn) {
try {
mysqli_close($conn);
} catch (Exception $e) {
echo $e->getMessage();
}
}
function find( $table = null, $id = null ) {
$database = open_database();
$found = null;
try {
if ($id) {
$sql = "SELECT * FROM" . $table . " WHERE id = " . $id;
$result = $database->query($sql);
if ($result->num_rows > 0) {
$found = $result->fetch_assoc();
}
}
else {
$sql = "SELECT * FROM " . $table;
$result = $database->query($sql);
if ($result->num_rows > 0) {
$found = $result->fetch_all(MYSQLI_ASSOC);
}
}
} catch (Exception $e) {
$_SESSION['message'] = $e->GetMessage();
$_SESSION['type'] = 'danger';
}
close_database($database);
return $found;
}
function find_all( $table ) {
return find($table);
}
function save($table = null, $data = null) {
$database = open_database();
$columns = null;
$values = null;
//print_r($data);
foreach ($data as $key => $value) {
$columns .= trim($key, "'") . ",";
$values .= "'$value',";
}
$columns = rtrim($columns, ',');
$values = rtrim($values, ',');
$sql = "INSERT INTO " . $table . "($columns)" . " VALUES " ($values);";
try {
$database->query($sql);
$_SESSION['message'] = 'Registro cadastrado com sucesso.';
$_SESSION['type'] = 'success';
} catch (Exception $e) {
$_SESSION['message'] = 'Nao foi possivel realizar a operacao.';
$_SESSION['type'] = 'danger';
}
close_database($database);
}
function update($table = null, $id = 0, $data = null) {
$database = open_database();
$items = null;
foreach ($data as $key => $value) {
$items .= trim($key, "'") . "='$value',";
}
$items = rtrim($items, ',');
$sql = "UPDATE " . $table;
$sql .= " SET $items";
$sql .= " WHERE id=" . $id . ";";
try {
$database->query($sql);
$_SESSION['message'] = 'Registro atualizado com sucesso.';
$_SESSION['type'] = 'success';
}
catch (Exception $e) {
$_SESSION['message'] = 'Nao foi possivel realizar a operacao.';
$_SESSION['type'] = 'danger';
}
close_database($database);
}
Sorry if it's not right idled.
I put an space on the code after the "FROM" at
$sql = "SELECT * FROM" . $table . " WHERE id = " . $id;
The error remains the same but know on line 36 that is:
if ($result->num_rows > 0) {
$found = $result->fetch_all(MYSQLI_ASSOC);
}
Turned the code on line 36 to:
$if (result = $database->query($sql)) {
The error disappeared, others problems not relative to this question happened.
Just in case this question isn't closed as off-topic -> typo generated, I'd better provide an acceptable answer.
Add a space between FROM and $table in your SELECT query.
Check for a non-false result from your query.
Your new code could look like this:
$sql = "SELECT * FROM `$table`";
if ($id) {
// assuming id has been properly sanitized
$sql .= " WHERE id=$id"; // concatenate with WHERE clause when appropriate
}
if ($result = $database->query($sql)) { // only continue if result isn't false
if ($result->num_rows) { // only continue if one or more row
$found = $result->fetch_all(MYSQLI_ASSOC); // fetch all regardless of 1 or more
}
}
$sql = "INSERT INTO " . $table . "($columns)" . " VALUES " ($values);"; has too many double quotes in it -- you can tell by how Stack Overflow highlights the characters. I could say that you could clean up the syntax, but it would ultimately be best to scrap your script and implement prepared statements.
I Have a json array with id type answer I'm merging id, type and answer together and put them in the id2 column, so why can't I get $answers value in output?
[{"id":"38","answer":[{"option":"3","text":"HIGH"}],"type":"a"},
{"id":"39","answer":[{"option":"3","text":"LOW"}],"type":"b"},
{"id":"40","answer":["Hello Word"],"type":"c"}]
This is my code:
<?php
$con=mysqli_connect("localhost","root","","arrayok");
mysqli_set_charset($con,"utf8");
// Check connection
if (mysqli_connect_errno()){
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$sql="SELECT `survey_answers`,us_id FROM `user_survey_start`";
if ($result=mysqli_query($con,$sql)){
while ($row = mysqli_fetch_row($result)){
$json = $row[0];
if(!is_null($json)){
$json = preg_replace("!\r?\n!", "", $json);
$jason_array = json_decode($json,true);
// id2
$id = array();
foreach ($jason_array as $data) {
if (array_key_exists('id', $data)) {
if (array_key_exists('type', $data)) {
if (array_key_exists('answer', $data)) {
foreach($data['answer'] as $ans){
$answers[] = isset($ans['text']) ? $ans['text'] : $ans;
}
$id[] = ' ID='.$data['id'].', TYPE='.$data['type'].', AWNSER='.$answers;
}
}
}
}
// lets check first your $types variable has value or not?
$ids= implode(',',$id); /// implode yes if you got values
$sql1="update user_survey_start set id2='$ids' where us_id=".$row[1];//run update sql
echo $sql1."<br>";
mysqli_query($con,$sql1);
}
}
}
mysqli_close($con);
?>
And this is my output:
update user_survey_start set id2=' ID=38, TYPE=a, AWNSER=Array,
and I got Notice: Array to string conversion in C:\wamp64\www\json\awnser.php on line 29
I want to have value of $answers
Solved By Myself
Because I Couldn't Put Awnser Directly To id[], I Taked Awnser Like This:
$id[] = ' ID='.$data['id'].', TYPE='.$data['type'];
$id[] = isset($ans['text']) ? ' AWNSER='.$ans['text'] : ' AWNSER='.$ans;
And This is My Code:
<?php
$con=mysqli_connect("localhost","root","","arrayok");
mysqli_set_charset($con,"utf8");
// Check connection
if (mysqli_connect_errno()){
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$sql="SELECT `survey_answers`,us_id FROM `user_survey_start`";
if ($result=mysqli_query($con,$sql)){
while ($row = mysqli_fetch_row($result)){
$json = $row[0];
if(!is_null($json)){
$json = preg_replace("!\r?\n!", "", $json);
$jason_array = json_decode($json,true);
// id2
$id = array();
foreach ($jason_array as $data) {
if (array_key_exists('id', $data)) {
if (array_key_exists('type', $data)) {
if (array_key_exists('answer', $data)) {
foreach($data['answer'] as $ans){
$id[] = ' ID='.$data['id'].', TYPE='.$data['type'];
$id[] = isset($ans['text']) ? ' AWNSER='.$ans['text'] : ' AWNSER='.$ans;
}
}
}
}
}
// lets check first your $types variable has value or not?
$ids= implode(',',$id); /// implode yes if you got values
$sql1="update user_survey_start set id2='$ids' where us_id=".$row[1];//run update sql
echo $sql1."<br>";
mysqli_query($con,$sql1);
}
}
}
mysqli_close($con);
?>
By using below code
$data = array();
$sql = "SELECT * FROM list";
$result = $db->query($sql);
if ($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
$data['title'] = $row['title'];
$data['name'] = $row['name'];
}
}
echo json_encode($data);
I got 1 result, I can get full result if I do $data[] = $row['title'], but I want to make the result like this
{'title' : ['title 1','title 2'], 'name':['John','Amy']}
You are overwriting the title in each loop iteration. You need to accumulate all the titles and then set it in your data array.
$data = array();
$sql = "SELECT title FROM mainlist";
$result = $db->query($sql);
$titles = array();
if ($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
$titles[] = $row['title'];
}
}
$data['title'] = $titles;
echo json_encode($data);
The easiest away is probably this:
$rows = array();
$sql = "SELECT * FROM list";
$result = $db->query($sql);
if ($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
$rows[] = $row;
}
}
echo json_encode($rows);
you could achieve by using group_concat on each of the columns in your query. That way you do not need to loop the result again and add column etc...
$sql = "SELECT group_concat(title) as title,group_concat(name) as name FROM list";
$result = $db->fetch(PDO::FETCH_ASSOC);
echo json_encode($result);
Try this:
$titles = array();
$names = array();
$sql = "SELECT title,name FROM mainlist";
$result = $db->query($sql);
if ($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
$titles[] = $row['title'];
$names[] = $row['name'];
}
}
echo json_encode(array("title" => $titles, "name" => $names));
UPDATE
Updated my code to let you manage an undefined number of columns as result
$out = array();
$sql = "SELECT * FROM wp_cineteca";
$result = $db->query($sql);
if ($result->num_rows > 0) {
while($row = $result->fetch_array()) {
$keys = array_keys($row);
for ($i = 0; $i < count($row); $i++) {
$out[$keys[$i]][] = $row[$i];
}
}
}
echo json_encode($out);
I want to Print some dynamically data from my function to passing string in function like this.
My Function
function mysql_funX_Repeter($query,$itemtemplet)
{
$this->mysql_funX_connect();
$result = mysql_query($query) or die("Repeter Query Error.");
if (!$result)
{
$message = 'ERROR:' . mysql_error();
return $message;
}
else
{
$newtempelt = str_replace("Eval", '$row', $itemtemplet);
while ($row = mysql_fetch_array($result))
{
echo $newtempelt;
}
}
}
Passing Value In Function
<ul>
<?php
$rptvalue="<li><a href=''>Eval['name']</a></li>";
$myFun->mysql_funX_Repeter("Select name from tbldemo",$rptvalue);
?>
</ul>
My Output
$row['name']
$row['name']
$row['name']
I want Output
raj
ram
prince
But it's Show me only variable but not show it's database value.
How to solve this...!!!
I would change a little bit your method.
function mysql_funX_Repeter($query,$itemtemplet)
{
$this->mysql_funX_connect();
$result = mysql_query($query) or die("Repeter Query Error.");
if (!$result)
{
$message = 'ERROR:' . mysql_error();
return $message;
}
else
{
while ($row = mysql_fetch_array($result))
{
echo "<li><a href=''>".$row[$itemtemplet]."</a></li>";
}
}
}
And of course, your method invokation:
<ul>
<?php
$rptvalue="name";
$myFun->mysql_funX_Repeter("Select ".$rptvalue." from tbldemo",$rptvalue);
?>
</ul>
And that's how I'd do it to avoid nasty evals. :)
function mysql_funX_Repeter($query,$itemtemplet)
{
$html='';
$this->mysql_funX_connect();
$result = mysql_query($query) or die("Repeter Query Error.");
$num_rows=mysql_num_rows($result);
if($num_rows){
while ($row = mysql_fetch_array($result))
{
$newtempelt = str_replace("Eval", $row['column_name'], $itemtemplet);
$html.="<li><a href=''>".$newtempelt."</a></li>";
}
}
return $html;
}
//calling method mysql_funX_Repeter
<ul>
<?php
$rptvalue="name";
echo $myFun->mysql_funX_Repeter("Select ".$rptvalue." from tbldemo",$rptvalue);
?>
</ul>
Developer My Problem Was Solved.
Thanks for Reply.
My Function
//This Repeter Function Created By Priyank Patel
function mysql_funX_Repeter($query,$itemtemplet)
{
$this->mysql_funX_connect();
$result = mysql_query($query) or die("Repeter Query Error.");
if (!$result)
{
$message = 'ERROR:' . mysql_error();
return $message;
}
else
{
preg_match_all('/{([^}]*)}/', $itemtemplet, $matches);
$select = '';
while($row = mysql_fetch_assoc($result)){
$aux = $itemtemplet;
for($i = 0; $i < count($matches[0]); $i++){
$aux = str_replace($matches[0][$i], $row[$matches[1][$i]],$aux);
}
$select .= $aux."\n";
}
mysql_close();
return $select;
}
}
Calling Style
<ul><?php $templet="<li>{name}</li>";echo $myFun->mysql_funX_Repeter("Select * from tbldemo",$templet);?></ul>
1) I want to save case 1 value to an array. Return this array and pass it to a function. The code become case 2, but no result come out, where is the problem?
2) In function display_urls, i want to echo both $url and $category. What should i do in IF condition or add another line of code?
function display_urls($url_array)
{
echo "";
if (is_array($url_array) && count($url_array)>0)
{
foreach ($url_array as $url)
{
echo "".$url."";
echo "".$category."";
}
}
echo "";
}
case 1: work fine
$result = oci_parse($conn, "select * from bookmark where username ='$username'");
if (!$result){ $err = oci_error(); exit; }
$r = oci_execute($result);
$i=0;
echo "";
while( $row = oci_fetch_array($result) ){
$i++;
echo "";
echo "".$row['USERNAME']."";
echo "".$row['BM_URL']."";
echo "".$row['CATEGORY']."";
echo "";
}
echo "";
case 2:
$url_array = array();
while( $row2 = oci_fetch_array($result, OCI_BOTH)){
$i++;
$url_array[$count] = $row[0];
}
return $url_array;
I think you probably want something like this:
function display_urls($url_array)
{
echo "";
if (is_array($url_array) && count($url_array)>0)
{
foreach ($url_array as $url)
{
echo "".$url['BM_URL']."";
echo "".$url['CATEGORY']."";
}
}
echo "";
}
$result = oci_parse($conn, "select * from bookmark where username ='$username'");
if (!$result){ $err = oci_error(); exit; }
$r = oci_execute($result);
$url_array = array();
while( $row = oci_fetch_array($result, OCI_ASSOC)){
$url_array[] = $row;
}
display_urls($url_array);
This will store all the information on the URLs in $url_array with a lookup by column name.