MySQL can't accept numbers variable - php

I'm facing this weird problem. The code below does not work. I only get 0 from Count. Expected output is 12 from my 12 rows.
echo $userid = $row['userid']; // output is 130
mysqli_query($con, "SELECT COUNT(*) AS count FROM client WHERE userid = $userid");
However if I replace $userid variable with integer, it will work:
echo $userid = $row['userid']; // output is 130
mysqli_query($con, "SELECT COUNT(*) AS count FROM client WHERE userid = 130");
UPDATE:
$row['userid'] value was taken from other column with the column type as int.
I need to use variable in query. Please help me and thanks in advance.
UPDATE:
So I test with the code below where I assign 130 to $test. It works.
$test = 130;
mysqli_query($con, "SELECT COUNT(*) AS count FROM client WHERE userid = $test");

I guess it has something to do with parameter being of string type, where you need an integer. Also 'count' is a reserved word which may cause an error, not handled by your code. Need to backtick that also.
I suggest to use prepared statements. It is safer, better practice, and hopefully will solve your issue. Example:
$userid = $row['userid'];
/* create a prepared statement */
if ($stmt = mysqli_prepare($link, "SELECT COUNT(*) AS `count` FROM client WHERE userid =?")) {
/* bind parameters for markers */
mysqli_stmt_bind_param($stmt, "i", $userid);
/* execute query */
mysqli_stmt_execute($stmt);
/* bind result variables */
mysqli_stmt_bind_result($stmt, $district);
/* fetch value */
mysqli_stmt_fetch($stmt);
Documentation:
http://php.net/manual/en/mysqli.prepare.php

This is pretty awkward:
echo $userid = $row['userid']; // output is 130
Go ahead and get rid of that echo statement.

Related

What can be problem in this line with PDO? [duplicate]

I have a mysql query that targets a single column in a single row
"SELECT some_col_name FROM table_name WHERE user=:user"
After I execute the statement $stmt->execute(); how do I get this single cell directly placed into a variable with no loops? In other words how to get
from $stmt->execute();
to $col_value = 100;
I tried the 2 below, but neither worked.. The column is number 4 in the original table, but I'm assuming since in my select statement I'm selecting it only, it should be 1 when I specify the parameter for fetchColumn.
$col_value = $stmt->fetchColumn();
$col_value = $stmt->fetchColumn(0);
As you can see, I'm trying to do it in as few lines as possible.
Are you sure it's returning any rows?
$stmt->fetchColumn()
is correct way to fetch a single value, so either you probably didn't bind the :user parameter or it simply returned no rows.
$sql='SELECT some_col_name FROM table_name WHERE user=?';
$sth=$pdo_dbh->prepare($sql);
$data=array($user);
$sth->execute($data);
$result=$sth->fetchColumn();
I'm not sure why so many people mess this up:
$stmt = $dbh->prepare("SELECT `column` FROM `table` WHERE `where`=:where");
$stmt->bindValue(':where', $MyWhere);
$stmt->execute();
$SingleVar = $stmt->fetchColumn();
Make sure that you are selecting a specific column in the query and not * or you will need to specify the column order number in fetchColumn(), example: $stmt->fetchColumn(2); That usually isn't a good idea because the columns in the database may be reorganized by, someone...
This will only work properly with unique 'wheres'; fetchColumn() will not return an array.
When you want to get the last insert you add the DESC Limit 1 to the sql statement.
$sql = "SELECT `some_col_name` FROM table_name\n"
. "ORDER BY `some_col_name` DESC\n"
. "LIMIT 1";
$stmt = $conn->prepare($sql);
$result = $stmt->execute();
$row = $stmt->fetch(PDO::FETCH_ASSOC);
//convert the array content to string and store in variable
$col = implode(" ", $row);
echo $col;
Have you prepared the statement first? (Before $stmt->execute())
$stmt = $db->prepare("SELECT some_col_name FROM table_name WHERE user=:user");
You could use this:
$stmt->fetch(PDO::FETCH_COLUMN, $number_of_column);

Using PHP variable in SQL query

I'm having some trouble using a variable declared in PHP with an SQL query. I have used the resources at How to include a PHP variable inside a MySQL insert statement but have had no luck with them. I realize this is prone to SQL injection and if someone wants to show me how to protect against that, I will gladly implement that. (I think by using mysql_real_escape_string but that may be deprecated?)
<?php
$q = 'Hospital_Name';
$query = "SELECT * FROM database.table WHERE field_name = 'hospital_name' AND value = '$q'";
$query_result = mysqli_query($conn, $query);
while ($row = mysqli_fetch_assoc($query_result)) {
echo $row['value'];
}
?>
I have tried switching '$q' with $q and that doesn't work. If I substitute the hospital name directly into the query, the SQL query and PHP output code works so I know that's not the problem unless for some reason it uses different logic with a variable when connecting to the database and executing the query.
Thank you in advance.
Edit: I'll go ahead and post more of my actual code instead of just the problem areas since unfortunately none of the answers provided have worked. I am trying to print out a "Case ID" that is the primary key tied to a patient. I am using a REDCap clinical database and their table structure is a little different than normal relational databases. My code is as follows:
<?php
$q = 'Hospital_Name';
$query = "SELECT * FROM database.table WHERE field_name = 'case_id' AND record in (SELECT distinct record FROM database.table WHERE field_name = 'hospital_name' AND value = '$q')";
$query_result = mysqli_query($conn, $query);
while ($row = mysqli_fetch_assoc($query_result)) {
echo $row['value'];
}
?>
I have tried substituting $q with '$q' and '".$q."' and none of those print out the case_id that I need. I also tried using the mysqli_stmt_* functions but they printed nothing but blank as well. Our server uses PHP version 5.3.3 if that is helpful.
Thanks again.
Do it like so
<?php
$q = 'mercy_west';
$query = "SELECT col1,col2,col3,col4 FROM database.table WHERE field_name = 'hospital_name' AND value = ?";
if($stmt = $db->query($query)){
$stmt->bind_param("s",$q); // s is for string, i for integer, number of these must match your ? marks in query. Then variable you're binding is the $q, Must match number of ? as well
$stmt->execute();
$stmt->bind_result($col1,$col2,$col3,$col4); // Can initialize these above with $col1 = "", but these bind what you're selecting. If you select 5 times, must have 5 variables, and they go in in order. select id,name, bind_result($id,name)
$stmt->store_result();
while($stmt->fetch()){ // fetch the results
echo $col1;
}
$stmt->close();
}
?>
Yes mysql_real_escape_string() is deprecated.
One solution, as hinted by answers like this one in that post you included a link to, is to use prepared statements. MySQLi and PDO both support binding parameters with prepared statements.
To continue using the mysqli_* functions, use:
mysqli_prepare() to get a prepared statement
mysqli_stmt_bind_param() to bind the parameter (e.g. for the WHERE condition value='$q')
mysqli_stmt_execute() to execute the statement
mysqli_stmt_bind_result() to send the output to a variable.
<?php
$q = 'Hospital_Name';
$query = "SELECT value FROM database.table WHERE field_name = 'hospital_name' AND value = ?";
$statement = mysqli_prepare($conn, $query);
//Bind parameter for $q; substituted for first ? in $query
//first parameter: 's' -> string
mysqli_stmt_bind_param($statement, 's', $q);
//execute the statement
mysqli_stmt_execute($statement);
//bind an output variable
mysqli_stmt_bind_result($stmt, $value);
while ( mysqli_stmt_fetch($stmt)) {
echo $value; //print the value from each returned row
}
If you consider using PDO, look at bindparam(). You will need to determine the parameters for the PDO constructor but then can use it to get prepared statements with the prepare() method.

Cannot echo result from mysql

All I need is to produce a row. I've looked at all the samples and I cannot for the life of me get the right information. Hence help is required please.
Connection to DB in the usual way. Here is my code for the query.
$sql = "SELECT * FROM table WHERE `u_password` = $pword AND `user` = $uname LIMIT 1";
$result = mysqli_query($mdb, $sql);
$row = mysqli_fetch_assoc($result);
//Then I try to retrieve say the user name....
echo $row['seeking'];
I've got a count in there and it produces a result of 1.
The error I get is
'Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result'
Help would be appreciated.
The error
Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result
Almost always means that the query failed for some reason, thus $result = mysqli_query returns FALSE rather than a mysql_result object so anything that then tries to use $result as an object will not work for obvious reasons.
The issue with your query is that text column data must be wrapped in quotes like this
$sql = "SELECT *
FROM table
WHERE `u_password` = '$pword' AND `user` = '$uname' LIMIT 1";
Your script is at risk of SQL Injection Attack
Have a look at what happened to Little Bobby Tables Even
if you are escaping inputs, its not safe!
You should use parameterized queries to avoid this.
$sql = "SELECT *
FROM table
WHERE `u_password` = ? AND `user` = ? LIMIT 1";
$stmt = mysqli_prepare($mdb, $sql);
// its also a good idea to check the staus of a prepare
// and show the error if it failed, at least while testing
if ( $stmt === FALSE ) {
echo mysqli_error($mdb);
exit;
}
$stmt->bind_param('ss', $pword, $uname );
$stmt->execute();
$result = $stmt->get_result();
$row = $result->fetch_assoc();
echo $row['seeking'];
You need to use prepared statements (in actuality you could get it to work by quoting your strings but prepared statements are much better). Like so:
$sql = "SELECT * FROM table WHERE `u_password` = ? AND `user` = ? LIMIT 1";
$stmt = mysqli_prepare($mdb, $sql);
$stmt->bind_param("ss",$pword,$uname);
if ($stmt->execute()) {
$result = $stmt->get_result();
$row = mysqli_fetch_assoc($result);
//Then I try to retrieve say the user name....
echo $row['seeking'];
} else { /* something went wrong */ }

PHP, MySQL statement results in ZERO rows

hope someone can help me.
i have a very simple prepared SELECT statment in PHP:
$query_select = ("SELECT * FROM companies where user_name = ? ");
$stmt = $mysqli->prepare($query_select);
$stmt->bind_param("s", $user_name);
$stmt->execute();
$count = $stmt->num_rows;
in companies table I have several rows with the $user_name i`m trying to query. But i still get 0 rows as a result.
The strange thing is that the non PREPARED version works:
$query = 'SELECT * FROM companies WHERE user_name="'.$user_name.'"';
$result = $mysqli->query($query);
$count= $result->num_rows;
echo "Aantal: ".$count;
So my question is, does anyone know why the prepared version returns ZERO and the non prepared version returns the correct number of rows?
Add this line to your code between execute and num_rows statement.
$stmt->store_result();
You have to store it before counting it.
For mysqli prepared statements, you must take an additional step: storing the result.
Try this:
$query_select = ("SELECT * FROM companies where user_name = ? ");
$stmt = $mysqli->prepare($query_select);
$stmt->bind_param("s", $user_name);
$stmt->execute();
$stmt->store_result(); // <-- new line
$count = $stmt->num_rows;
May be you need to bind the result:
/* bind result variables */
$stmt->bind_result($district);
Full example here

How to solve Illegal string offset ['id'] in php?

I'm sorry if this is a duplicate question but I don't know how to solve my problem. Every time I try to correct my error I fail. My code is:
if (isset($_GET["comment"])) {$id = $_GET["comment"];}
$query = "SELECT * FROM posts WHERE id = {$id['$id']};";
$get_comment = mysqli_query($con, $query);
Can anybody correct the code to not show an error anymore and tell me what did I wrong?
Try this:
$id = isset($_GET['comment']) ? $_GET['comment'] : 0;
$query = "SELECT * FROM `posts` WHERE `id` = " . intval($id);
The use of intval will protect you from SQL injection in this particular case. Ideally, you should learn PDO as it is extremely powerful and makes prepared statements much easier to handle to prevent all injections.
An example using PDO might look like:
$id = isset($_GET['comment']) ? $_GET['comment'] : 0;
$query = $pdo->prepare("SELECT * FROM `posts` WHERE `id` = :id");
$query->execute(array("id"=>$id));
$result = $query->fetch(PDO::FETCH_ASSOC); // for a single row
// $results = $query->fetchAll(PDO::FETCH_ASSOC); // for multiple rows
var_dump($result);
First of all you should prevent injestion.
if (isset($_GET["comment"])){
$id = (int)$_GET["comment"];
}
Notice, $id contanis int.
$query = "SELECT * FROM posts WHERE id = {$id}";
Assuming your $id is an integer and you only want to make the query if it is set, here's how you could do it using prepared statements, which protect you from MYSQL injection attacks:
if (isset($_GET["comment"])) {
$id = $_GET["comment"];
$stmt = mysqli_prepare($con, "SELECT * FROM posts WHERE id = ?");
mysqli_stmt_bind_param($stmt, 'i', $id);
mysqli_stmt_execute($stmt);
mysqli_stmt_bind_result($stmt, $get_comment);
while (mysqli_stmt_fetch($stmt)) {
// use $get_comment
}
mysqli_stmt_close($stmt);
}
Most of these functions return a boolean indicating whether they were successful or not, so you might want to check their return values.
This approach looks a lot more heavy duty and is arguably overkill for a simple case of a statement containing a single integer but it's a good practice to get into.
You might want to look at the object-oriented style of mysqli which you might find a little cleaner-looking, or alternatively consider using PDO.

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