how do i parse an array element within double-quotes - php

Is there more elegant way of escaping array elements:
mysql_query("SELECT * from mytable WHERE name = '".$someArray['somename']."'");
Something more like
mysql_query("SELECT * from mytable WHERE name = '$myname'");
So avoiding the annoying '". ."'"


You can use:
mysql_query("SELECT * from mytable WHERE name = '{$someArray['somename']}'");


You can actually do:
mysql_query("SELECT * from mytable WHERE name = '{$foo[bar]}'");
But I would advise against it or any other parsed string, it's slower than concatenation and is much harder to read in some IDEs.


$myname = $someArray['somename'];
mysql_query("SELECT * from mytable WHERE name = '$myname'");

Related

Using SQL LIKE with variable PHP

I can't use the SQL LIKE with a variable
$sql = "SELECT * FROM chat WHERE name LIKE 'Motherboard' "; //This works
But if a use a variable it doesn't work:
$sql = "SELECT * FROM chat WHERE name LIKE '%'+$variable+'%' ";
//or
$sql = "SELECT * FROM chat WHERE name LIKE '$variable' ";
How can I fix it?

In PHP you concatenate strings with a dot .
$sql = "SELECT * FROM chat WHERE name LIKE '" . $variable . "' ";

You also can use variables within double quotes:
$sql = "SELECT * FROM chat WHERE name LIKE '$variable' ";
or
$sql = "SELECT * FROM chat WHERE name LIKE '{$variable}' ";
Keep in mind that double quotes work slower then single.
In your case it would be
$sql = "SELECT * FROM chat WHERE name LIKE '%{$variable}%' ";

Using PHP variables dynamically in MySQL statements

I am trying to execute the query below.
$condition = "WHERE emp_id = '$emp_id'";
$myquery = "SELECT * FROM emp_table".$condition;
I expect my query to be like this, but dynamically:
$myquery = "SELECT * FROM emp_table WHERE emp_id = '$emp_id'";
Is there anyway to make SQL statements dynamically through variables in php..?

It should be
$condition = "WHERE emp_id = '$emp_id'";
$myquery = "SELECT * FROM emp_table ".$condition;
you forgot to put $ on myquery

You should never build queries dynamically like that. Correct way to do it is to use prepared statements.
In your case it'll be like that
$statement = $pdo->prepare("SELECT * FROM emp_table WHERE emp_id = :emp_id");
$statement->execute(array(
':emp_id' => $emp_id
));
$rows = $statements->fetchAll(PDO::FETCH_ASSOC);

Have you tried this? -
$condition = "WHERE emp_id = '" . $emp_id . "'";
$myquery = "SELECT * FROM emp_table " . $condition;
PHP does not expand variables within single-quote strings.
echo "$name"; //works
echo '$name'; //does not work
Take a look at this: Single quotes or double quotes for variable concatenation?

Short Code for PHP string with variable

I have a query string that contains a variable like this
$field_name = 'features';
$value = '5';
$query = "SELECT * FROM Table WHERE $field_name\_tid = '$value'";
My goal is to print out the $query like this SELECT * FROM Table WHERE features_tid = '5';
I put \_ there hoping it would work as escape character, but it didn't work. Is there any way to achieve this without use methods like ". $field_name ." and modifying original variable value?

yes:
$query = "SELECT * FROM Table WHERE {$field_name}_tid = '$value'";

You can use:
$query = "SELECT * FROM Table WHERE {$field_name}_tid = '$value'";

How to select with a binary field ? (php,mysql)

Try to select use "where" clause in a mysql statement:
e.g.
Table: X with a ID column which is BINARY data type. Then save in a variable in php
$aid = $row["id"];
How do i use this variable later when I try to select from table
$where = "where `ID` = '$aid'";
$query = "SELECT * FROM X ".$where;
Return 0 row.
Does anyone know why?

Answering my own question.
Just figured out:
$where = "where HEX(ID) = 'bin2hex($aid)'";
$query = "SELECT * FROM X ".$where;
Does anyone know better solution?

Try below :
add BINARY in where clause.
$where = "where BINARY ID = '$aid'";
$query = "SELECT * FROM X ".$where;

Simple mysql query not working

I have this very simple function:
function getCatName($id){
$sql = "SELECT * FROM biznet_category WHERE ID ='".$id."';";
$res = mysql_query ($sql) or die (mysql_error ());
$row = mysql_fetch_assoc ($res);
$name = $row["Name"];
return $name;
}
So with this function I should be able to get the category name, but it doesn't work with the parameter. If I put 8 or 9, the categoryname is displayed correctly.
The id is also passed on like it should, when I print it out, it shows 8 or 9.
I know the solution is quite simple, I just don't see it.

To fix remove the quotes and check the column name for case id or ID. Since the query string is in double quotes you don't have to use the . join
$sql = "SELECT * FROM biznet_category WHERE ID = $id";
You can use curly brackets which I find easier to read
$sql = "SELECT * FROM biznet_category WHERE ID = {$id}";
If you were querying a string rather than an integer you can simply do
$sql = "SELECT * FROM biznet_category WHERE ID = '{$id}'";

$sql = "SELECT * FROM biznet_category WHERE ID ='".$id."';";
To
$sql = "SELECT * FROM biznet_category WHERE ID = ".$id;

Try this
$sql = "SELECT * FROM biznet_category WHERE ID = ".$id;

Is the column name ID spelt correctly?

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