Here, I am getting error like,
Notice: Trying to get property of non-object
in last two lines while fetching record.
what does it say?
My code:
$Id = $_REQUEST['id'];
$sql = "Select * From ".CHANNEL_MASTER."
Where sam_status = '".ACTIVE_STATUS."' And user_id = '".$_SESSION['user_id']."' And sam_id = '".$Id."'";
$db->query($sql);
$row = $db->fetch_object(MYSQL_FETCH_SINGLE);
$siteID = array_search($row->sam_site_id, $site_id_array);
$ebay_token = $row->sam_ebay_token;
You need to store query result into a variable then fetch data from it.
So instead of
$db->query($sql);
$row = $db->fetch_object(MYSQL_FETCH_SINGLE);
use
$result=$db->query($sql);// store query result into $result
$row = $result->fetch_object(MYSQL_FETCH_SINGLE);// fetch data from $result
$Id = $_REQUEST['id'];
$sql = "Select * From ".CHANNEL_MASTER."
Where sam_status = '".ACTIVE_STATUS."' And user_id = '".$_SESSION['user_id']."' And sam_id = '".$Id."'";
$result = $db->query($sql);
$row = $result->fetch_object(MYSQL_FETCH_SINGLE);
$siteID = array_search($row->sam_site_id, $site_id_array);
$ebay_token = $row->sam_ebay_token;
Related
i'm trying to sum a column name "total". and i want to display the total sorting by id. if user A login he can see total booking in his account.
I keep get the error:
"Notice: Array to string conversion in Array."
can someone help me? I want to echo the total in input form.
this is my php code:
<?php
include ('connect.php');
$sql = "SELECT * FROM penjaga WHERE p_username = '".$_SESSION['username']."'";
$result = mysqli_query($conn,$sql);
$row = mysqli_fetch_assoc($result);
$id = $row['p_id'];
$sql2 = "SELECT SUM(total) as total FROM sitter_kucing WHERE sitter_fk = '$id'";
$row2 = mysql_fetch_array($sql);
$sum = $row['total'];
?>
Try this,
$sql2 = "SELECT SUM(total) as total FROM sitter_kucing WHERE sitter_fk = '$id'";
$result2 = mysql_query($sql2) or die(mysql_error());
$row2 = mysql_fetch_array($result2) or die(mysql_error());
$sum = $row['total'];
i got it! thanks this is my code
this is the code:
<?php
include ('connect.php');
$sql8 = "SELECT * FROM penjaga WHERE p_username = '".$_SESSION['username']."'";
$result8 = mysqli_query($conn,$sql8);
$row8 = mysqli_fetch_assoc($result8);
$id = $row8['p_id'];
$sql9 = "SELECT SUM(total) as total FROM sitter_kucing WHERE sitter_fk = '$id'";
$result9 = mysqli_query($conn,$sql9);
$row9 = mysqli_fetch_array($result9);
$sum = $row9['total'];
?>
for ($i=$start; $i<$start+$scale && $i < $total_record; $i++)
{
$sql = "select * from memo where num = ?";
$stmh = $pdo->prepare($sql);
//mysql_data_seek($result, $i);
$row = mysql_fetch_array($result);
$sql2 = "select * from phptest.memo order by num desc";
$stmh2 = $pdo->query($sql2);
$stmh2->execute();
//$row = $stmh2->fetch(PDO::FETCH_ASSOC);
$row = $stmh->fetchColumn($i-1);
$memo_id = $row['id'];
$memo_num = $row['num'];
$memo_date = $row['regist_day'];
$memo_nick = $row['nick'];
$memo_content = $row['content'];
Hi guys i want fetch single row data by using PDO method instead of $row = $stmh2->fetch(PDO::FETCH_ASSOC); like mysqli_data_seek($result,$i);. What should i do?
$sql2 = "select * from phptest.memo order by num desc";
$stmh2 = $pdo->query($sql2);
$stmh2->execute();
while($r = $stmh2->fetch()) {$row[] = $r;}
Now $row[$i] should be getting you the same results as mysqli_data_seek($result,$i) would have.
IE: $row[0] would return the first row.
Can someone help me what's the problem with this code? Im trying to store the fetched data to an array and i want to based on the values of that array. Im getting an error of Array to string conversion. The datatype value of an array is string
Here's the code.
$sql3 ="SELECT DISTINCT subj_descr FROM subj_enrolled WHERE enroll_ref = '$ref'";
$results = mysqli_query($con, $sql3);
$data = array();
while($row = mysqli_fetch_array($results)){
$data[] = array($row['subj_descr']);
}
$sql ="SELECT * FROM notification WHERE subj_descr IN ({implode(',', $data})";
$result = mysqli_query($con, $sql);
$count = mysqli_num_rows($result);
Remove array inside your while loop:
$sql3 ="SELECT DISTINCT subj_descr FROM subj_enrolled WHERE enroll_ref = '$ref'";
$results = mysqli_query($con, $sql3);
$data = array();
while($row = mysqli_fetch_array($results)){
$data[] = $row['subj_descr'];
}
$sql ="SELECT * FROM notification WHERE subj_descr IN ({implode(',', $data})";
$result = mysqli_query($con, $sql);
$count = mysqli_num_rows($result);
create a new variable and implode in it.
Try this
$implodeAray = implode(",", $data);
$sql ="SELECT * FROM notification WHERE subj_descr IN ($implodeAray)";
You are creating a multidimensional array and so change this statement
$data[] = array($row['subj_descr']);
to
$data[] = $row['subj_descr'];
As SQL IN statement always used a single dimensional array so also make change in query where clause.
I have changed all, please try below code:
<?php
$sql3 ="SELECT DISTINCT subj_descr FROM subj_enrolled WHERE enroll_ref = '$ref'";
$results = mysqli_query($con, $sql3);
$data = array();
while($row = mysqli_fetch_array($results)){
$data[] = $row['subj_descr'];
}
$dataStr = implode(',', $data);
$sql ="SELECT * FROM notification WHERE subj_descr IN (".$dataStr.")";
$result = mysqli_query($con, $sql);
$count = mysqli_num_rows($result);
?>
You stored the element of your array inside another array while looping.
Do this:
$sql3 ='SELECT DISTINCT subj_descr
FROM subj_enrolled
WHERE enroll_ref = "$ref"';
$results = mysqli_query($con, $sql3);
$data = array();
while($row = mysqli_fetch_array($results)){
//Your error was here
//Each elements is escaped for security reasons
$data[] = mysqli_escape_string($con,$row['subj_descr']);
}
//This implodes and puts a single quote around each element
$dataIn= '\'' . implode( '\', \'', $data ) . '\'';
$sql ="SELECT * FROM notification
WHERE subj_descr IN ($dataIn)";
$result = mysqli_query($con, $sql);
$count = mysqli_num_rows($result);
I am trying to store a mysql value into a php variable. I have the following query which I know works. However, I the value for $count is always 0. Can someone explain what I need to do to get the count value? The count should be the count of x's w here name_x=.$id.
$query = "SELECT COUNT(name_x) FROM Status where name_x=.$id.";
$result = mysql_query($query);
$count = $result;
Is first letter in table name is really capital. Please check it first.
or Try :
$query = "SELECT COUNT(*) as totalno FROM Status where name_x=".$id;
$result = mysql_query($query);
while($data=mysql_fetch_array($result)){
$count = $data['totalno'];
}
echo $count;
$query = "SELECT COUNT(*) FROM `Status` where `name_x`= $id";
$result = mysql_query($query);
$row = mysql_fetch_row($result);
$count = $row[0];
please try it
$query = "SELECT COUNT(*) FROM Status where name_x=$id";
$result = mysql_query($query);
$count = mysql_result($result, 0);
You are missing single quotes around $id. Should be
name_x = '" . $id . "'";
I am build a data structure using multidimensional associative arrays. Can I update that data structure into a mysql table field and then read it back again?
Here is an example of what I am trying to do:
$result = mysql_query("select * FROM color") or die(mysql_error());
$colors = "";
while($colorrec = mysql_fetch_array($result)){
$colors[$colorrec['ID']][0] = $colorrec['Description'];
$colors[$colorrec['ID']][1] = $colorrec['HexCode'];
}
If I now do:
mysql_query("UPDATE tempfile SET ColorInfo = '".$colors."' WHERE ID = '".tempID."'");
Can I then do:
$result = mysql_query("select * from tempfile WHERE ID = '".tempID."'");
$temprec = mysql_fetch_array($result);
$colors = $temprec['ColorInfo'];
You could serialize it to hold it's data type, then unserialize it when you fetch it back.
$result = mysql_query("select * FROM color") or die(mysql_error());
$colors = array();
while($colorrec = mysql_fetch_array($result)){
$colors[$colorrec['ID']] = array($colorrec['Description'], $colorrec['HexCode']);
}
mysql_query("UPDATE tempfile SET ColorInfo = '".serialize($colors)."' WHERE ID = '".$tempID."'");
$result = mysql_query("select * from tempfile WHERE ID = '".$tempID."'");
$temprec = mysql_fetch_array($result);
$colors = unserialize($temprec['ColorInfo']);