php select query help

willy ze great · Apr 8, 2016

database hairetrieve data za mtu anayetaka kuona vitu alivyonunua kutoka table ya manunuzi
lakini nikiedit hapo "where username='john'" inaretrieve data zote za john napata but now napata Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result
<?php
$host="localhost";
$user="root";
$password="";
$database="online";
$username=$_SESSION['username'];
$link= mysqli_connect($host, $user, $password,$database);
$query="select * from manunuzi where username=$username order by date desc";
$result= mysqli_query($link, $query);
echo "<table border='1' bgcolor='WHITE'>";
echo "<caption>YOUR BIDDING TABLE</caption>";
echo "<tr><th>PRODUCT</th><th>DATE</th>
<th>QUANTITY</th><th>PRICE</th><th>METHOD</th></tr>";
while ($row =mysqli_fetch_array($result)) {
echo "<tr>";
echo "<td>".$row['product']."</td>";
echo "<td>".$row['date']."</td>";
echo "<td>".$row['qty']."</td>";
echo "<td>".$row['price']."</td>";
echo "<td>".$row['method']."</td>";
echo "</tr>";
}
?>

MastaKiraka · Apr 8, 2016

Hapo kwenye $query jaribu kicheki hizo semi colon

willy ze great · Apr 8, 2016

kirakapacha said:
Hapo kwenye $query jaribu kicheki hizo semi colon

mkuu hio semi colon inafunga sql query na ya pili inaterminate hio statement i think iko sawa anyway nimetoa ila bado napata hio warning

MastaKiraka · Apr 8, 2016

Hapo kwenye $query weka username = '$username'. i.e add quotes kwenye hiyo variable username

Kang · Apr 8, 2016

willy ze great said:
database hairetrieve data za mtu anayetaka kuona vitu alivyonunua kutoka table ya manunuzi
lakini nikiedit hapo "where username='john'" inaretrieve data zote za john napata but now napata Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result
<?php
$host="localhost";
$user="root";
$password="";
$database="online";
$username=$_SESSION['username'];
$link= mysqli_connect($host, $user, $password,$database);

/* check connection */
if (mysqli_connect_errno()) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}

$query="select * from manunuzi where username=$username order by date desc";
$result= mysqli_query($link, $query);

if($result===FALSE){
printf("Query: " . $query);
printf(" Failed");
exit();
}

echo "<table border='1' bgcolor='WHITE'>";
echo "<caption>YOUR BIDDING TABLE</caption>";
echo "<tr><th>PRODUCT</th><th>DATE</th>
<th>QUANTITY</th><th>PRICE</th><th>METHOD</th></tr>";
while ($row =mysqli_fetch_array($result)) {
echo "<tr>";
echo "<td>".$row['product']."</td>";
echo "<td>".$row['date']."</td>";
echo "<td>".$row['qty']."</td>";
echo "<td>".$row['price']."</td>";
echo "<td>".$row['method']."</td>";
echo "</tr>";
}
?>

Validate results zako kabla haujazitumia, utajua tatizo liko wapi. (Nimeedit code hapo cheki maandishi mekundu.)

Pia jaribu kusetup debugger ambayo itakuwezesha kupitia code yako step by step wakati inaexecute na kuangalia values zako cheki Xdebug youtube.

Mwisho unavyounga username kwenye SQL query hizo kuna hatari ya kuwa na security issue ya SQL Injection.
How can I prevent SQL-injection in PHP?

frankgalos · Apr 8, 2016

willy ze great said:
database hairetrieve data za mtu anayetaka kuona vitu alivyonunua kutoka table ya manunuzi
lakini nikiedit hapo "where username='john'" inaretrieve data zote za john napata but now napata Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result
<?php
$host="localhost";
$user="root";
$password="";
$database="online";
$username=$_SESSION['username'];
$link= mysqli_connect($host, $user, $password,$database);
$query="select * from manunuzi where username=$username order by date desc";
$result= mysqli_query($link, $query);
echo "<table border='1' bgcolor='WHITE'>";
echo "<caption>YOUR BIDDING TABLE</caption>";
echo "<tr><th>PRODUCT</th><th>DATE</th>
<th>QUANTITY</th><th>PRICE</th><th>METHOD</th></tr>";
while ($row =mysqli_fetch_array($result)) {
echo "<tr>";
echo "<td>".$row['product']."</td>";
echo "<td>".$row['date']."</td>";
echo "<td>".$row['qty']."</td>";
echo "<td>".$row['price']."</td>";
echo "<td>".$row['method']."</td>";
echo "</tr>";
}
?>

jaribu kutumia OOP way

$host="localhost";
$user="root";
$password="";
$database="online";
$username=$_SESSION['username'];
$link= new mysqli($host, $user, $password,$database);

$query= $link->query("select * from manunuzi where username=$username order by date desc");
if($query->num_rows > 0){
while ($row = $query->fetch_array($result)) {
echo "<tr>";
echo "<td>".$row['product']."</td>";
echo "<td>".$row['date']."</td>";
echo "<td>".$row['qty']."</td>";
echo "<td>".$row['price']."</td>";
echo "<td>".$row['method']."</td>";
echo "</tr>";
}
}

frankgalos · Apr 8, 2016

willy ze great said:
database hairetrieve data za mtu anayetaka kuona vitu alivyonunua kutoka table ya manunuzi
lakini nikiedit hapo "where username='john'" inaretrieve data zote za john napata but now napata Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result
<?php
$host="localhost";
$user="root";
$password="";
$database="online";
$username=$_SESSION['username'];
$link= mysqli_connect($host, $user, $password,$database);
$query="select * from manunuzi where username=$username order by date desc";
$result= mysqli_query($link, $query);
echo "<table border='1' bgcolor='WHITE'>";
echo "<caption>YOUR BIDDING TABLE</caption>";
echo "<tr><th>PRODUCT</th><th>DATE</th>
<th>QUANTITY</th><th>PRICE</th><th>METHOD</th></tr>";
while ($row =mysqli_fetch_array($result)) {
echo "<tr>";
echo "<td>".$row['product']."</td>";
echo "<td>".$row['date']."</td>";
echo "<td>".$row['qty']."</td>";
echo "<td>".$row['price']."</td>";
echo "<td>".$row['method']."</td>";
echo "</tr>";
}
?>

pia jaribu kuweka quotation mark katika variable ya $username it will work

php select query help

willy ze great

JF-Expert Member

MastaKiraka

JF-Expert Member

willy ze great

JF-Expert Member

MastaKiraka

JF-Expert Member

Kang

JF-Expert Member

frankgalos

JF-Expert Member

frankgalos

JF-Expert Member

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