I need to read a text file, query the database table with that name, and store that table's data in another table. So far I have written this code but I don't know why it's not working.
foreach ($lindb as $namedb) {
$query = "SELECT * FROM ntable WHERE name =" .$namedb. "";
$result = mysql_query($query);
while ($r = mysql_fetch_array($result)) {
$query = "INSERT INTO ndtable (name,details,address,login,country) VALUES (\"".$r["name"]."\", \"".$r["details"]."\", \"".$r["address"]."\", \"".$r["login"]."\", \"".$r["country"]."\")";
mysql_query($query);
}
}
You don't have quotes around $namedb
ie. SELECT * FROM ntable WHERE name =" .$namedb. ""; should be SELECT * FROM ntable WHERE name ='" .$namedb. "'";
I suggest a SELECT INTO would be the better choice... and please post the error so we are able to help...
Related
I want to display the table participantes with the columns sorteo, nombre and fecha.
The user info is on another table sellify_users (usern column).
I want to display only that user data using:
SELECT * FROM participantes WHERE nombre = 'usern'
But usern is not in the same table, so if possible I want to call the sellify_users to get the usern data.
<?php
$user = 'database_user';
$password = 'database_pass';
$database="database_name";
mysql_connect(localhost,$user, $password);
#mysql_select_db($database) or die( "Unable to select database");
echo $query = "SELECT * FROM participantes WHERE nombre='usern'";
$result = mysql_query($query);
mysql_close();
?>
If you meant that a user has a record in the table sellify_users and you need to find the usern from it in order to use it in the next query:
$query = "SELECT * FROM participantes WHERE nombre='usern'";
Then all you need to do is to run a query for the table sellify_users first and get the value usern from it, store it in a variable and then use that in your second query. Something like:
$query1 = "SELECT * FROM sellify_users";
$result1 = mysqli_query($con, $query1);
$row1 = mysqli_fetch_assoc($result1);
$usern = $row1['usern'];
$query2 = "SELECT * FROM participantes WHERE nombre='usern'";
$result2 = mysqli_query($con, $query2);
while($row2 = mysqli_fetch_assoc($result2)){
echo $row2['ColumnNameHere1'];
echo $row2['ColumnNameHere2'];
}
Notice: I've passed $con which denotes a connection variable, i.e.
you should read about mysqli or PDO and if there's anything you do not understand, feel free to shoot a query.
EDIT:
If by any chance you are trying to use the last inserted record, you should look for mysqli_insert_id($con); and use that instead.
The following works with no problem at all, when the photoId is directly on the statement and not a variable.
$img_query = mysqli_query($con, 'SELECT * FROM imgs WHERE photoid = "103"') or die(mysqli_error($con));
but the following just won't work with no error, what might be causing this not to select.
$imageid = '103';
$img_query = mysqli_query($con, 'SELECT * FROM imgs WHERE photoid = "$imageid"') or die(mysqli_error($con));
$img_row = mysqli_fetch_array($img_query);
echo $img_row['img'];
This is inside a while loop.
while($row = mysqli_fetch_array($somequery)){
$imageid = $row['photoid'];
$img_query = mysqli_query($con, 'SELECT * FROM imgs WHERE photoid = "$imageid"') or die(mysqli_error($con));
$img_row = mysqli_fetch_array($img_query);
echo $img_row['img'];
}
Thanks.
in php a ' and a " are very different and the query syntax is double quote around the query and single quote around variables.. although I would recommend you look at using parameters on your query instead of just putting a variable directly into the query
Per my recommendation you should change your query to this:
$imageid = '103';
$query = $con->prepare("SELECT * FROM imgs WHERE photoid = ?");
$query->bind_param('sssd', $imageid);
$query->execute();
this is just the nuts and bolts of it... if you want more information about the connection.. error handling and everything else read the DOCS
there is a big difference between ' and " in php
Differences
change your query to be
$img_query = mysqli_query($con, "SELECT * FROM imgs WHERE photoid = '$imageid'") or die(mysqli_error($con));
and it should work.
Hello I have 3 fields on input form which are set via POST method to external php
$id=$_POST['id'];
$nombre=$_POST['nombre'];
$cedula=$_POST['cedula'];
where I would like to make a search option depending on which field have data inside it or if a user put data in all 3 or in only 2 fields to search from the input fields which are not NULL fields in the same table where there is a result.
my sql query is something like that $sql = "SELECT * FROM users WHERE userID = $id AND nombre = $nombre AND cedula = $cedula) ";
obviosly which is not working, what should I do to make it work. Do I need to change only the query or I need to put something before it to check first what is not NULL. Thanks
Firstly, your SQL statement should be updated to have enclosed ' (commas) around string values.
So, modify it to:
$sql = "SELECT * FROM users WHERE userID = '$id' AND nombre = '$nombre' AND pass = '$pass'";
// ----------------------------------------^---^--------------^-------^------------^-----^
Second thing is that you should search a field only when it has a value otherwise, it of no use.
So, your modified PHP code should be:
$sql = "SELECT * FROM users WHERE 1 ";
if (!empty($id)) {
$sql .= " AND userID = '$id' ";
}
if (!empty($nombre)) {
$sql .= " AND nombre= '$nombre' ";
}
if (!empty($pass)) {
$sql .= " AND pass= '$pass' ";
}
And your Database will be searched for the fields only if they have data filled in the form.
Try to add quote:
$sql = "SELECT * FROM users WHERE userID = ".$id." AND nombre = ".$nombre." AND pass = '".$pass."' ";
Yes, you will need to put a check before which will ignore the fields which are null.
Also, you would need to put the $variable inside single quotes ' if they are VARCHAR or CHAR types.
How make mysql search defined just by what is written in html form, by user, and if some form box is stayed empty, mysql should ignore it. For example:
$sql = "SELECT * FROM catalog WHERE name= '".$name."' AND publisher = '".$publisher."' ";
mysql_query($sql);
This query will display all rows where name and publisher are together. Now, what if user insert just name, and left publisher box empty. The idea is that php/mysql ignore empty form box, and display every row with inserted name. But it will not do that because $publisher will be undefined, and error emerges. How to tell musql to ignore $publisher? More generally, the question is: how to generate query that make searching defined by certain criteria if they exists, and if they don't how to just ignore it?
You can build up the sql programmatically. I am assuming you have escaped the values properly.
$sql = "SELECT * FROM catalog";
$wheres = array();
if (!empty($name)) {
$wheres[] = " name = '$name'";
}
if (!empty($publisher)) {
$wheres[] = " publisher = '$publisher'";
}
if (count($wheres)) {
$sql .= " WHERE " . implode (' AND ', $wheres);
}
//RUN SQL
Also have a read through this, you are using a deprecated mysql library.
This will allow either the name or the publisher to be NULL.
<?php
$sql = "SELECT * FROM catalog WHERE (name= '".$name."' OR name IS NULL) AND (publisher = '".$publisher."' OR publisher IS NULL)";
mysql_query($sql);
Try like
$my_var = " ";
if($publisher) //if(!empty($publisher))
$my_var = " AND publisher = '".$publisher."' ";
$sql = "SELECT * FROM catalog WHERE name= '".$name."' ".$my_var;
if the publisher is empty then you need to pass the NULL value and PLZ note that it is a bad practise.It will causes many sql injection issues.Try to put validations for the things
I am trying to get four different values from my database. The session variable username and usernameto are working, but I want to get 4 different values -- two each from username and usernameto:
<?php
session_start(); // startsession
$Username=$_SESSION['UserID'];
$Usernameto= $_SESSION['UserTO'];
$db = mysql_connect("at-web2.xxxxxx", "yyyyy", "xxxxxxx");
mysql_select_db("db_xxxxxx",$db);
$result1 = mysql_query("SELECT user_lon and user_lat FROM table1 WHERE id = '$Usernameto'");
$result2 = mysql_query("SELECT user_lon and user_lat FROM table1 WHERE id = '$Username'");
$myrow1 = mysql_fetch_row($result1);
$myrow2 = mysql_fetch_row($result2);
while($myrow1)
{
$_Mylon=$myrow1[0];
$_Mylat=$myrow1[1];
}
while($myrow2)
{
$_Mylon2=$myrow2[0];
$_Mylat2=$myrow2[1];
}
?>
Edit - just realized that you didn't tell us what wasn't working about the code you provided. Are you getting an error message or are you not getting the correct data back? You still should fix your query, but we'll need some more information to know what's wrong.
Your query statements shouldn't have "and" between the select parameters, so it should be:
Edit 2 - I just noticed that you had a while loop that you don't need, try this:
$result1 = mysql_query("SELECT user_lon, user_lat FROM table1 WHERE id = '$Usernameto'");
$result2 = mysql_query("SELECT user_lon, user_lat FROM table1 WHERE id = '$Username'");
$myrow1 = mysql_fetch_row($result1);
$myrow2 = mysql_fetch_row($result2);
if (isset($myrow1)) {
$_Mylon=$myrow1[0];
$_Mylat=$myrow1[1];
}
if (isset($myrow2)) {
$_Mylon2=$myrow2[0];
$_Mylat2=$myrow2[1];
}
An example from the php manual echoing an html table
I don't know if you can derive what you need from this?
More specific: You can use:
$line = mysql_fetch_array($result, MYSQL_ASSOC);