PHP Interview Questions Answers

PHP Interview Question - 1 : -

PHP Interview Answer - 1 : -

PHP Interview Question - 2 : -

PHP Interview Answer - 2 : -

PHP Interview Question - 3 : -

PHP Interview Answer - 3 : -

1) - Create a full backup of your database: shell> mysqldump tab=/path/to/some/dir opt db_name
Or: shell> mysqlhotcopy db_name /path/to/some/dir
The full backup file is just a set of SQL statements, so restoring it is very easy:

shell> mysql "."Executed";

PHP Interview Question - 4 : -

PHP Interview Answer - 4 : -

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PHP Interview Answer - 5 : -

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PHP Interview Answer - 6 : -

var myValue =9.8;
if(parseInt(myValue)== myValue)
alert('Integer');
else
alert('Not an integer');

Tools used for drawing ER diagrams.
Case Studio
Smart Draw

PHP Interview Question - 7 : -

PHP Interview Answer - 7 : -

There are two possible ways to bridge PHP and Java: you can either integrate PHP into a Java Servlet environment, which is the more stable and efficient solution, or integrate Java support into PHP. The former is provided by a SAPI module that interfaces with the Servlet server, the latter by this Java extension.
The Java extension provides a simple and effective means for creating and invoking methods on Java objects from PHP. The JVM is created using JNI, and everything runs in-process.

Example Code:

getProperty('java.version') . ''; echo 'Java vendor=' . $system->getProperty('java.vendor') . ''; echo 'OS=' . $system->getProperty('os.name') . ' ' . $system->getProperty('os.version') . ' on ' . $system->getProperty('os.arch') . ' '; // java.util.Date example $formatter = new Java('java.text.SimpleDateFormat', "EEEE, MMMM dd, yyyy 'at' h:mm:ss a zzzz"); echo $formatter->format(new Java('java.util.Date')); ?>

The behaviour of these functions is affected by settings in php.ini.
Table 1. Java configuration options
Name
Default
Changeable
java.class.path
NULL
PHP_INI_ALL
Name Default Changeable
java.home
NULL
PHP_INI_ALL
java.library.path
NULL
PHP_INI_ALL
java.library
JAVALIB
PHP_INI_ALL

PHP Interview Question - 8 : -

PHP Interview Answer - 8 : -

PHP Interview Question - 9 : -

PHP Interview Answer - 9 : -

There is at least 3 ways to logon to a remote server:
Use ssh or telnet if you concern with security
You can also use rlogin to logon to a remote server.

Please give a regular expression (preferably Perl/PREG style), which can be used to identify the URL from within a HTML link tag.
Try this: /href="([^"]*)"/i

PHP Interview Question - 10 : -

PHP Interview Answer - 10 : -

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PHP Interview Answer - 11 : -

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PHP Interview Answer - 12 : -

PHP Interview Question - 13 : -

PHP Interview Answer - 13 : -

PHP does not require (or support) explicit type definition in variable declaration; a variable's type is determined by the context in which that variable is used. That is to say, if you assign a string value to variable $var, $var becomes a string. If you then assign an integer value to $var, it becomes an integer.

An example of PHP's automatic type conversion is the addition operator '+'. If any of the operands is a float, then all operands are evaluated as floats, and the result will be a float. Otherwise, the operands will be interpreted as integers, and the result will also be an integer. Note that this does NOT change the types of the operands themselves; the only change is in how the operands are evaluated.

$foo += 2; // $foo is now an integer (2)
$foo = $foo + 1.3; // $foo is now a float (3.3)
$foo = 5 + "10 Little Piggies"; // $foo is integer (15)
$foo = 5 + "10 Small Pigs"; // $foo is integer (15)

If the last two examples above seem odd, see String conversion to numbers.
If you wish to change the type of a variable, see settype().
If you would like to test any of the examples in this section, you can use the var_dump() function.
Note: The behavior of an automatic conversion to array is currently undefined.

Since PHP (for historical reasons) supports indexing into strings via offsets using the same syntax as array indexing, the example above leads to a problem: should $a become an array with its first element being "f", or should "f" become the first character of the string $a? The current versions of PHP interpret the second assignment as a string offset identification, so $a becomes "f", the result of this automatic conversion however should be considered undefined. PHP 4 introduced the new curly bracket syntax to access characters in string, use this syntax instead of the one presented above:

PHP Interview Question - 14 : -

PHP Interview Answer - 14 : -

PHP Interview Question - 15 : -

PHP Interview Answer - 15 : -

If you want to create a table, you can run the CREATE TABLE statement as shown in the following sample script:

<?php
include "mysql_connection.php";

$sql = "CREATE TABLE Tech_links ("
. " id INTEGER NOT NULL"
. ", url VARCHAR(80) NOT NULL"
. ", notes VARCHAR(1024)"
. ", counts INTEGER"
. ", time TIMESTAMP DEFAULT sysdate()"
. ")";
if (mysql_query($sql, $con)) {
print("Table Tech_links created.\n");
} else {
print("Table creation failed.\n");
}

mysql_close($con);
?>

Remember that mysql_query() returns TRUE/FALSE on CREATE statements. If you run this script, you will get something like this:
Table Tech_links created.