Failing to recognize fetch_assoc method - php
I'm trying to upload an excel file to my site and save the data in my database, however i'm failing to do so and getting: Fatal error: Call to undefined method mysqli::fetch_assoc() ... But i'm not sure how to handle it, and i haven't found a question related on SO, any help?
function getSchedule($filepath,$con,$filename){
require_once 'excel/PHPExcel/IOFactory.php';
$objPHPExcel = PHPExcel_IOFactory::load($filepath);
foreach ($objPHPExcel->getWorksheetIterator() as $worksheet) {
$worksheetTitle = $worksheet->getTitle();
$highestRow = $worksheet->getHighestRow(); // e.g. 10
$highestColumn = $worksheet->getHighestColumn(); // e.g 'F'
$highestColumnIndex = PHPExcel_Cell::columnIndexFromString($highestColumn);
list($location, $date) = explode('-', $filename, 2);
$LastChange = date('d/m/Y h:i:s');
$Status='Open';
$servername = "localhost";
$username = "root";
$password = "Js";
$dbname = "jr";
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT * FROM Schedule";
$conn->query($sql);
// output data of each row
while($row = $conn->fetch_assoc()) {
$sql1="DELETE FROM `schedule` WHERE " . $row["Date"]. "='$date'";
$result = mysqli_query($conn,$sql1);
}
$conn->close();
for ($row = 3; $row <= $highestRow; ++ $row) {
$sql="INSERT INTO `schedule` (`Status`,`LastChange`, `Location`,`Date`,`AFNumber`,`Name`,`01-IN`, `01-OUT`, `02-IN`, `02-OUT`, `03-IN`, `03-OUT`, `04-IN`, `04-OUT`, `05-IN`, `05-OUT`, `06-IN`, `06-OUT`, `07-IN`, `07-OUT`, `08-IN`, `08-OUT`, `09-IN`, `09-OUT`, `10-IN`, `10-OUT`, `11-IN`, `11-OUT`, `12-IN`, `12-OUT`, `13-IN`, `13-OUT`, `14-IN`, `14-OUT`, `15-IN`, `15-OUT`, `16-IN`, `16-OUT`, `17-IN`, `17-OUT`, `18-IN`, `18-OUT`, `19-IN`, `19-OUT`, `20-IN`, `20-OUT`, `21-IN`, `21-OUT`, `22-IN`, `22-OUT`, `23-IN`, `23-OUT`, `24-IN`, `24-OUT`, `25-IN`, `25-OUT`, `26-IN`, `26-OUT`, `27-IN`, `27-OUT`, `28-IN`, `28-OUT`, `29-IN`, `29-OUT`, `30-IN`, `30-OUT`, `31-IN`, `31-OUT`) VALUES ('".$Status."', '".$LastChange."','".$location."','".$date."',";
for ($col = 0; $col < ($highestColumnIndex -1); ++ $col) {
$cell = $worksheet->getCellByColumnAndRow($col, $row);
$val = $cell->getValue();
if($col==($highestColumnIndex -2)){
$sql.="'$val'";
}else{
$sql.="'$val', ";}
}
echo "Index:".$highestColumnIndex."<br>";
if($highestColumnIndex < 63){
$temp = 63 - $highestColumnIndex;
for($i = 1;$i <= $temp; $i++){
if($i == $temp){
$sql.=",''";
} else{
$sql.=", '',";
}
}
}
$sql .=")";
if ($con->query($sql) === TRUE) {
} else {
echo "<br><br>Error: " . $sql . "<br>" . $con->error;
}
}//End For Each Row
}//End For Each Worksheet
}//End getHours Function
From php.net:
array mysqli_result::fetch_assoc ( void )
This means that you should provide it with a result of a query, not a connection. Change your code to this and it should work
$results = $conn->query($sql); //assign query to a variable and get mysqli_result in return
while($row = $results->fetch_assoc()) { //use that in the while loop
You are overwriting the connection object $conn.Use
$result = $conn->query($sql);
while($row = $result->fetch_assoc()) {
Try this:
$sql = "SELECT * FROM Schedule";
$result = $conn->query($sql);
// output data of each row
while($row = $result->fetch_assoc()) {
Related
Multiple for-loops from one $result = $conn->query($query)
I have imported data into a table and I am now accessing that data with PHP using the code as follows, <?php require_once 'connect.php'; $query = "SELECT * FROM JunkData"; $result = $conn->query($query); if(!$result) die("Fatal Error"); $rows = $result->num_rows; for ($name = 0; $name < $rows; ++$name) { $row = $result->fetch_array(MYSQLI_ASSOC); echo htmlspecialchars($row['Name']) . '<br/>'; } $result->close(); $conn->close(); This works! I am really just curious why adding a second for-loop does not work, unless I declare $result again? <?php require_once 'connect.php'; $query = "SELECT * FROM JunkData"; $result = $conn->query($query); if(!$result) die("Fatal Error"); $rows = $result->num_rows; for ($name = 0; $name < $rows; ++$name) { $row = $result->fetch_array(MYSQLI_ASSOC); echo htmlspecialchars($row['Name']) . '<br/>'; } for ($number = 0; $number < $rows; ++$number) { $row = $result->fetch_array(MYSQLI_ASSOC); echo htmlspecialchars($row['Number']) . 'Flag<br/>'; } $result->close(); $conn->close(); Doesn't work, although 'Flag' is printed an appropriate number of times. Whereas if I declare $result again. <?php require_once 'connect.php'; $query = "SELECT * FROM JunkData"; $result = $conn->query($query); if(!$result) die("Fatal Error"); $rows = $result->num_rows; for ($name = 0; $name < $rows; ++$name) { $row = $result->fetch_array(MYSQLI_ASSOC); echo htmlspecialchars($row['Name']) . '<br/>'; } $result = $conn->query($query); for ($number = 0; $number < $rows; ++$number) { $row = $result->fetch_array(MYSQLI_ASSOC); echo htmlspecialchars($row['Number']) . '<br/>'; } $result->close(); $conn->close(); The code does work. I have tried unsetting a few variables with unset($row) etc and I did notice if I remove the line, $row = $result->fetch_array(MYSQLI_ASSOC); from the second for loop, it will print the last value in the Number column as many times as the loop will run. I hope that is understandable. I am wondering what is happening in the code that I need to re-declare $result if I want to run a second for loop against it.
The standard solution I would recommend is to do fetch_all() Instead of: $rows = $result->num_rows; for ($name = 0; $name < $rows; ++$name) { $row = $result->fetch_array(MYSQLI_ASSOC); Do $rows = $result->fetch_all(); foreach ($rows as $row) { ... } // then you can re-loop same array of $rows foreach ($rows as $row) { ... }
There is an internal pointer when you call fetch_* function. In your first for loop, you send the pointer to the end of the result set. So, the next fetch will return nothing. If you run $result->data_seek(0) you will reset this pointer and can reuse: for ($name = 0; $name < $rows; ++$name) { $row = $result->fetch_array(MYSQLI_ASSOC); echo htmlspecialchars($row['Name']) . '<br/>'; } $result->data_seek(0); //<---- REPLACE HERE for ($number = 0; $number < $rows; ++$number) { $row = $result->fetch_array(MYSQLI_ASSOC); echo htmlspecialchars($row['Number']) . '<br/>'; } Of course, usually there is no need to loop the same result set twice, you may need to rethink your logic and loop only once.
MySQL Data to JSON Array with inserting a string
This is my Code for now but it only works until the for loop in the mySQLResultsetToJSON(). <?php $servername = "127.0.0.1"; $username = "root"; $password = "123456"; $table = "ticketing"; $link = new mysqli($servername, $username, $password, $table); if ($link->connect_error) { die("Connection failed: " . $link->connect_error); } echo "Connected successfully"; $result = $link->query("SELECT * from Ticket"); print_r($result); $final_result = mySQLResultsetToJSON($result); print_r($final_result); $link->close(); function mySQLResultsetToJSON($resultSet) { for($i = 0; sizeof($resultSet); $i++) { $rows = array(); while($r = mysqli_fetch_assoc($resultSet[$i])) { $rows[] = $r; } $jsonResult[$i] = json_encode(array('Results' => $rows)); } print_r($jsonResult); return $jsonResult; } ?> Thank you! Thomas
echo "mysql data<br />"; $result = $link->query("SELECT * from users"); print_r($result->fetch_object()); echo "<br />"; echo "in json<br />"; $res = ['Results' => $result->fetch_object() ]; echo json_encode($res); $link->close();
User like $result = $link->query("SELECT * from Ticket"); $rows = array(); while($r = mysqli_fetch_assoc($result)) { $rows[] = $r; } print "<pre>"; print_r(json_encode(array('Results' =>$rows))); $link->close();
Query to return connecting flights given source and destination
So I have a table called "Flights" that stores the source and destination of a flight. Say you have A -> B and B -> C in a table, and you give the source as 'A' and destination as 'C', the query should return the path between these two A - > B, B -> C. I have tried this query: $result = $mysqli -> query( "SELECT * FROM Flights a JOIN Flights b ON a.Destination = b.Source AND '{$source}' = a.Source AND '{$destination}' = b.Destination"); As a side note: I'm using mysqli in PHP. I have done this query in the mysql console and it returns everything I need. But if I ran this query in PHP on the example table I gave it would return B -> C only. Any ideas where I went wrong? I think it could have to do with the PHP string formatting. <?php if(isset($_POST['source'])){ $source = $_POST['source']; } if(isset($_POST['destination'])){ $destination = $_POST['destination']; } $servername = "localhost"; $username = "root"; $password = ""; $dbname = "Airport Tracking"; $table ="Flights"; $flag = 0; // Create connection $mysqli = new mysqli($servername, $username, $password, $dbname); if ($mysqli->connect_error) { die("Connection failed: " . $conn->connect_error); } if(isset($_POST['source']) && isset($_POST['destination'])){ $columns= $mysqli -> query("SHOW COLUMNS FROM $table"); $sql = "SELECT * FROM Flights a JOIN Flights b ON a.Destination = b.Source AND {$source} = a.Source AND {$destination} = b.Destination"; echo "$sql"; $result = $mysqli -> query($sql); if($result == FALSE){ echo "<br><b>Incorrect input</b>"; $flag = 1; } else if($result->num_rows == 0){ echo "<br><b>Returned no results</b>"; $flag = 1; } } $array = array(); $i = 0; if(isset($_POST['source']) && $flag == 0){ // Display results echo "<table>"; echo "<thead><tr>"; while ($row = $columns -> fetch_array(MYSQLI_BOTH)) { echo "<th>" .$row['Field']. "</th>"; $array[] = $row['Field']; } echo "</tr></thead>"; while ($row = $result -> fetch_array(MYSQLI_BOTH)) { echo "<tr>"; while ($i < sizeof($array)) { echo "<td>" .utf8_encode($row[$array[$i]]). "</td>"; $i++; } echo "</tr>"; $i = 0; } echo "</table>"; } $mysqli->close(); ?> That is my PHP code. I apologize because I know indenting is incorrect.
Why Getting only 1 array instead of many arrays?
I am a completely newbie in programming php I would like to make this code below return many arrays(to flash as3), however I only receive one array.Can anyone please pinpoint what is my mistake here? thanks. $data_array = ""; $i = 0; //if(isset($_POST['myrequest']) && $_POST['myrequest'] == "get_characters") //{ $sql = mysqli_query($conn, "SELECT * FROM ns_users ORDER BY Char_id"); while($row = mysqli_fetch_array($sql)) { $i++; $fb_name = $row["Username"]; $fb_id = $row["Fb_id"]; $fb_at = $row["Access_token"]; $fb_sig = $row["Fb_sig"]; $char_id = $row["Char_id"]; if($i == 1) { $data_array .= "$fb_name|$fb_id|$fb_at|$fb_sig|$char_id"; } else { $data_array .= "(||)$fb_name|$fb_id|$fb_at|$fb_sig|$char_id"; } echo "returnStr=$data_array"; exit(); }
When you write your exit insight your loop you stop executing your program and you get only one record. You should set the echo and exit after your while loop. $data_array = ""; $i = 0; $sql = mysqli_query($conn, "SELECT * FROM ns_users ORDER BY Char_id"); while($row = mysqli_fetch_array($sql)) { $i++; $fb_name = $row["Username"]; $fb_id = $row["Fb_id"]; $fb_at = $row["Access_token"]; $fb_sig = $row["Fb_sig"]; $char_id = $row["Char_id"]; if($i == 1) { $data_array .= "$fb_name|$fb_id|$fb_at|$fb_sig|$char_id"; } else { $data_array .= "(||)$fb_name|$fb_id|$fb_at|$fb_sig|$char_id"; } } echo "returnStr=$data_array"; exit();
Those two last line of your should be outside of your loop: $data_array = ""; $i = 0; //if(isset($_POST['myrequest']) && $_POST['myrequest'] == "get_characters") //{ $sql = mysqli_query($conn, "SELECT * FROM ns_users ORDER BY Char_id"); while($row = mysqli_fetch_array($sql)) { $i++; $fb_name = $row["Username"]; $fb_id = $row["Fb_id"]; $fb_at = $row["Access_token"]; $fb_sig = $row["Fb_sig"]; $char_id = $row["Char_id"]; if($i == 1) { $data_array .= "$fb_name|$fb_id|$fb_at|$fb_sig|$char_id"; } else { $data_array .= "(||)$fb_name|$fb_id|$fb_at|$fb_sig|$char_id"; } } echo "returnStr=$data_array"; exit();
If you would name the columns that you want in the SELECT then it's much simpler. Make sure to use MYSQLI_ASSOC in the fetch: $sql = mysqli_query($conn, "SELECT Username, Fb_id, Access_token, Fb_sig, Char_id FROM ns_users ORDER BY Char_id"); while($row = mysqli_fetch_array($sql, MYSQLI_ASSOC)) { $data_array[] = implode('|', $row); } echo "returnStr=" . implode('(||)', $data_array); exit();
How do I output certain index values from a foreach array?
This is the structure in the database: items |itemLink ---------------------- Kill Bill|Kill Bill link Preman |Preman link This is the code: $db = new PDO("mysql:host=$hostname;dbname=$database", $username, $password); $items = 'SELECT items FROM menus'; $itemLink = 'SELECT itemLink FROM menus'; $itemQuery = $db->query($items); $linkQuery = $db->query($itemLink); $fetchItem = $itemQuery->fetchAll(PDO::FETCH_ASSOC); $fetchLink = $linkQuery->fetchAll(PDO::FETCH_ASSOC); $merged = array_merge($fetchItem,$fetchLink); foreach($merged as $entry) { foreach( $entry as $key => $value ) { } } From the above code, how do I output only the items' datas?
Using the example above you could then do something like this to answer you question $result = mysql_query('Select * from names'); while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) { echo $row["FirstName"] . " " . $row["LastName"] . "<br>"; } mysql_close($conn); ?>
I would use something like this, not two arrays for something that could be one query. I have shown three methods, using var_dump or print_r will show how each works. $conn = mysql_connect($hostname, $username, $password); if (!$conn) { die('Could not connect to MySQL: ' . mysqli_connect_error()); } $db_selected = mysql_select_db('sample', $conn); if (!$db_selected) { die("Can\t use db : ' . mysql_error()"); } $result = mysql_query('Select * from names'); while ($row = mysql_fetch_array($result, MYSQL_NUM)) { print_r($row); } $result = mysql_query('Select * from names'); while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) { print_r($row); } $result = mysql_query('Select * from names '); while ($row = mysql_fetch_array($result, MYSQL_BOTH)) { print_r($row); } mysql_close($conn);