Please help me with this.
I have a php code which search first for the records that will be needed to inserted to another table
here is my code:
//search for split values (capacitors)
$capacitance =mysql_query("SELECT itemno, wwpn, SUBSTR(val, 1, LENGTH(val) / 2) as capacitor,
SUBSTR(val, LENGTH(val) / 2+1) as capasitance
FROM bom_csv WHERE boardnumber ='$board' and bom_csv.qty<>'' and bom_csv.qty !='qty';");
while($row =mysql_fetch_array($capacitance))
{
echo "<pre>";
echo $row['itemno'];
echo $row['capacitor'];
echo $row['capasitance'];
echo $row['wwpn'];
echo "</pre>";
$capacitor = $row['capacitor'];
$capacity =$row['capasitance'];
$adi_pn=$row['adi_pn'];
$itemno=$row['itemno'];
//insert into via update
$update =#mysql_query ("UPDATE bom_crunching SET capacitor ='$capacitor', capacitance ='$capacity' WHERE boardmodel ='$board' and adi_pn ='$adi_pn'");
if ($update)
{
echo "OKAY!";
}
else
{
echo "NOT OKAY!";
}
}
While executing it, I am not getting any errors. However, when I look at my query browser it doesn't have any data inserted.
The result I want is to insert all the records in the table via update statement because it has a default value of null.
Thanks in advance!
First step, change the line:
$update =#mysql_query ("UPDATE bom_crunching SET capacitor ='$capacitor', capacitance ='$capacity' WHERE boardmodel ='$board' and adi_pn ='$adi_pn'");
to:
$update =mysql_query ("UPDATE bom_crunching SET capacitor ='$capacitor', capacitance ='$capacity' WHERE boardmodel ='$board' and adi_pn ='$adi_pn'") or die( mysql_error() );
So NOW you will be informed what is the problem, than we can help you better.
;)
Related
I'm trying to set up an update query to add the last entry with the new entry but my new entry keeps doubling.
I keep looking for error but everything seems ok in not sure why the value of sum keeps doubling.
$sum ='1';
$sql = "update table set old = old +'$sum' where id='1'";
$query=mysqli_query($con,$sql);
if ($con->query($sql) === TRUE) {
echo '<script type="text/javascript">alert("A ok");</script>';
} else {
echo "Bigo Problem: " . $con->error;
}
Try This
$sum =1;
$sql = "update table set old = old +'$sum' where id=1";
if ($con->query($sql) == TRUE) {
echo '<script type="text/javascript">
alert ("A ok");
</script>';
} else {
echo "Bigo Problem: " . $con->error;
}
This is the reason for doubling sum because query exected twice by your code.
$query=mysqli_query($con,$sql);
if ($con->query($sql) === TRUE)
Remove one line and code will work fine.
issue with below two lines.
$query=mysqli_query($con,$sql);
if ($con->query($sql) === TRUE) {
Above two line make two entry.
The problem is that mysqli_query($con,$sql) executes your query....and then $con->query($sql) also executes your query.
So you are running the same query twice from two different commands.
You can just remove $query=mysqli_query($con,$sql); - it isn't needed.
First I'm hitting on a wall here and I really could use your help. I coded the database so I have it all up and working plus all the data inside. I worked the HTML and the CSS media print query and I have it how I want it to look exactly. All I have to do now is:
for every row of the mysql select table I have to fill every specific input form
of the html page I made and print it
Can someone give me a hint of how I can do that?
Assuming you want to connect to your database and simply fetch the id you can do the following.
Ensure you change my_host, my_user, my-password, my_databse,my_tablewith your configuration settings. Then if you want to fetch anything else thanid` just change it to the column name you are looking for.
Be aware we are using PHP here.
// Open Connection
$con = #mysqli_connect('my_host', 'my_user', 'my-password', 'my_databse');
if (!$con) {
echo "Error: " . mysqli_connect_error();
exit();
}
// Some Query
$sql = 'SELECT * FROM my_table';
$query = mysqli_query($con, $sql);
while ($row = mysqli_fetch_array($query))
{
echo $row['id'];
}
// Close connection
mysqli_close ($con);
Check this link to get a in-depth explanation.
You can do this with two pages. One page gives you the overview and the other page gives you a print preview of your invoice.
The overview:
// DB select stuff here
while ($row = $result->fetch(PDO::FETCH_ASSOC)) {
echo "<tr>\n";
echo " <td>".htmlspecialchars($row['what'])."</td>\n";
echo " <td>".htmlspecialchars($row['ever'])."</td>\n";
echo " <td>Detail</td>\n";
echo "</tr>\n";
}
The detail page:
$sql = 'SELECT your, columns FROM tab WHERE id = ?';
$stmt = $db->prepare($sql);
$stmt->execute(array($_GET['id']));
$row = $stmt->fetch(PDO::FETCH_ASSOC);
if (!$row) {
echo "There is no data for the given Id\n";
return;
}
echo "What: ".htmlspecialchars($row['what'])."<br />\n";
echo "Ever: ".htmlspecialchars($row['ever'])."<br />\n";
I have a variable of the logged in user ($steamid) that I need to use to select and echo specific fields from the database. I am using the following code, but it is working incorrectly. All database info is correct, the tables, columns, and variables are not misspelled. Not sure what I'm doing wrong.
<?php
$con=mysqli_connect("private","private","private","private");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$sql = "SELECT `bananas` FROM `es_player` WHERE `steamid` = '$steamID'";
if ($result=mysqli_query($con,$sql))
{
// Get field information for all fields
while ($fieldinfo=mysqli_fetch_field($result))
{
printf("bananas: %n",$fieldinfo->bananas);
}
// Free result set
mysqli_free_result($result);
}
mysqli_close($con);
?>
No errors are shown, it simply returns "bananas:" with nothing after it. I feel like I didn't to it correctly, does anyone know what I might've done wrong? Here is a screenshot of my database table so you know what it looks like http://puu.sh/gCY3d/983b738458.png.
Try this:
$query = Mysqli_Query($con, "SELECT * FROM `es_player` WHERE `steamid`='$steamID'") or die(mysql_error());
if( ! mysqli_num_rows($query) )
{
echo 'No results found.';
}
else
{
$bananas_array = mysqli_fetch_assoc($query);
$bananas = $bananas_array['bananas'];
echo 'Number of bananas: '. $bananas;
}
If this doesn't work, there is a problem with STEAM_ID format. You could try triming the IDs to be JUST a number, and add the STEAM_x:x: to it later.
I'm attempting to extract some data from a database and echo each result. The code below is code that I took from a textbook and then tried to modify to fit my own website that is hosted locally. I cannot see where I'm going wrong, no error messages are shown, just a blank screen when I run the scrip.
<?php #script 9.4 view top 5 recipients
// This script exctracts data from db and then displays each record in a table
DEFINE('SYSPATH','FOO');
require '../application/config/database.php';
require 'mysqli_connect.php';
$q = "SELECT alert_recipient as NAME
FROM alert
LIMIT 5;
";
$r = mysqli_query($dbc,$q);
// $dbc database connection comes from required mysqli_connect.php
if($r)
{
while($row = mysqli_fetch_array($r, MYSQLI_ASSOC)) {
echo $row['name'];
}
}
else {
echo "<p>ERROR</p>".mysqli_error($dbc);
}
?>
The code looks okay except for your echo $row['name'];, note that you are selecting NAME, uppercase.
Change your echo statement to be:
echo $row['NAME'];
because field names quoted within $row array are case sensitive.
(Can't comment yet)
Maybe the script works but there is no results to display. Check your database.
Here is my code:
$campagin_id = $_SESSION['campagin_id_for_camp'];
$query = "SELECT * FROM survey_result where campagin_id = ".$campagin_id;
$conn=mysql_connect($dbconfig['db_hostname'],$dbconfig['db_username'],$dbconfig['db_password']) or die(mysql_error());
mysql_select_db($dbconfig['db_name'],$conn);
$exec_query =mysql_query($query) or die(mysql_error());
$row=mysql_fetch_array($exec_query);
echo "<br> row = ".$row;
while ($row=mysql_fetch_array($exec_query)){
echo "I am In";
}
The Problem is that I am not getting anything in $row I cant get into the while loop, nothing shows up when I try to echo the value of $row, No error Nothing. Can you help me to find a problem in my code ?
Ps : The database is their. I have checked for the query for the corresponding value of $campagin_id. and also when i tried to echo $exec_query it echoed this : Resource id #8
PPS : The database have more than 7 record for each id so it doesn't matter if I call mysql_fetch_array($exec_query) more than once before going in to the while loop. and for the $campagin_id in the session their are many records present in the database.
You have written $row=mysql_fetch_array($exec_query) and then you are echoing something. and you are using the same in while.
Instead of:
$row=mysql_fetch_array($exec_query);
echo "<br> row = ".$row;
while ($row=mysql_fetch_array($exec_query)){
echo "I am In";
}
Use this (as per my knowledge you should not use $row=mysql_fetch_array() once you have used before while):
while ($row=mysql_fetch_array($exec_query)){
echo "I am In";
}
If the query returns Resource id #8 then that means it was successful - ie there were no errors. There were probably no rows returned by that query, so no rows in your table that match the given campagin_id.
You are also calling mysql_fetch_array() twice separately, you shouldn't do that because your while loop will skip the first row because calling this moves the pointer in the result set forward by one.
Also you can't echo an array as you are trying to, if you want to see the contents of an array use print_r() or var_dump().
I suggest adding some code to handle no rows found:
if($exec_query && mysql_num_rows($exec_query) > 0)
{
while ($row=mysql_fetch_array($exec_query)){
echo "Row: " . print_r($row, true);
}
}
else
{
echo 'None found';
}
Try this code.
<?
$campagin_id = $_SESSION['campagin_id_for_camp'];
$query = "SELECT * FROM survey_result where campagin_id = ".$campagin_id;
mysql_connect($dbconfig['db_hostname'],$dbconfig['db_username'],$dbconfig['db_password']) or die(mysql_error());
mysql_select_db($dbconfig['db_name']);
$exec_query =mysql_query($query) or die(mysql_error());
while ($row=mysql_fetch_assoc($exec_query)) {
echo "<br/> row = <pre>".print_r($row)."</pre><br/>";
}
?>