[Solved] You have an error in your SQL syntax…?

You will have to check if there is no starting and trailing comma’s in the $columns or $values variables. Plus, just to be sure, put appropriate quotes around the columns and values individually. public function insert($data, $table) { $columns = “”; $values = “”; foreach ($data as $column=>$value) { $columns .= “`” . $column . … Read more

[Solved] MS SQL query not working on MySQL [closed]

SELECT ppmap_d.*, p_prob.*, ppmap_h.*, p_probgroup.*, p_prod.*, ppmap_d.prob_id AS probid, ppmap_d.map_id AS mapid, ppmap_h.pg_name AS probgname, ppmap_h.m_id AS modelid FROM p_prob INNER JOIN p_prod ON ppmap_d.prob_id = p_probgroup.prob_id INNER JOIN ppmap_h ON ppmap_h.map_id = ppmap_d.map_id AND ppmap_h.pg_name = p_probgroup.pg_name INNER JOIN ppmap_d ON p_prod.m_id = ppmap_h.m_id INNER JOIN p_probgroup ON p_prob.prob_id = p_probgroup.prob_id; 4 solved MS … Read more

[Solved] PHP script to update mySQL database

Your sql is wrong. Apart from the gaping wide open SQL injection attack vulnerability, you’re generating bad sql. e.g. consider submitting “Fred” as the first name: $First_Name2 = “Fred”; $query = “UPDATE people SET Fred = First_name WHERE ….”; now you’re telling the db to update a field name “Fred” to the value in the … Read more

[Solved] Auto refresh PHP script backend [closed]

To refresh a page without reloading, you have to load the contents of the page through ajax. You could do this as follows: index.php <html> <head> <script type=”text/javascript”> $(document).on(‘ready’, function(){ setInterval(function() { $.ajax({ type: “GET”, url: “ajax_refresh.php”, success: function(result) { $(‘body’).html($result); } }); }, 3000); }); </script> </head> <body> <div class=”header”> Header of website here … Read more

[Solved] Simplify SQL query for me please [closed]

I am not sure if I got your question all clear, this query below will give you the latest employee record if there are multiple user records – SELECT * FROM employmentrecords WHERE id IN(SELECT MAX(id) FROM employmentrecords WHERE ((date_end >=’2017-08-22′ OR date_end IS NULL OR (date_end <=’2017-08-22′ AND date_end >=’2017-08-08′)) AND date_hired <=’2017-08-22′) GROUP … Read more

[Solved] A Parse error with the PHP mysql Class function of insert method

This line of code isn’t closed: $this->query(“INSERT INTO testing_Rand (number1, // etc and you are missing a ; after this line: $newRand.=implode(‘,’, $varNum) If I get it, you meant to write this instead: public function insert_SQL($varNum ) { $this->query(“INSERT INTO testing_Rand (number1, number2, number3, number4, number5, number6, number7, number8, number9, number10) VALUES (“.implode(‘,’, $varNum).”)”); } … Read more

[Solved] Why mysql_num_row less than condition doesn’t work?

Use this: <?php $sql=mysql_query(“SELECT ID FROM emp WHERE user=”$login_session””); // its currently zero $row=mysql_num_rows($sql); $b = 3; $tot=$b – $row; echo “$tot”; if($row < $b) { echo “good”; } else { echo “wrong”; } ?> You have been trying to output variables, which is not defined. You code was: Ouput undefined variables Assign something to … Read more

[Solved] How to store and display an image in MySQL database

Well, you gave a different name for the file in the input file tag in your form. Change this <input type=”file” name=”image” ><br> to this <input type=”file” name=”file” ><br> and it should work. Let me know if it doesn’t. You may also consider putting the contents in upload_file.php in an if-conditional to prevent it from … Read more

[Solved] mySQL data type for form elements

1) you will get string as the value of radio so it will be varchar type 2) for check boxes any one can have multiple values so you need to create a separate table(working_project) for this values( data type will be same as radio) and another table for mapping(user_working_project) user_working_project table will containing user id … Read more

[Solved] Relation to many and get without this

In SQL, this type of query needs what is known as an EXCEPTION JOIN. Some RDBMSs actually implement this as a separate type (such as DB2), while others need to use a workaround. In your case, it amounts to (in SQL): SELECT User.* FROM User LEFT JOIN UserHouse ON UserHouse.id_user = User.id WHERE UserHouse.id_user IS … Read more