PHP insert statement is only inserting 1 time - php

Question

im using php to insert into the review table.ive given the variables $email, $starcount, $bookid fixed values for now just to test the file. the $res query checks to see if there is a row with that book id and email in it. if theres not The $sql query inserts it, and then the $nex query loops through taking any starcount columns where the book column = $book.
if i change the the email at the top of the file it should insert into the new info database and pull out the new and existing starcount, but it does not post, it just returns the already existing starcount. i dont understand why its not working .... im using the array to return to my file.
<?php
mysql_connect("localhost","root","");
mysql_select_db("FunReads");
$email = "sd";
$starcount = "2";
$bookid = "5";
$res = mysql_query("SELECT * FROM Review WHERE book_id='$bookid' AND user_email='$email'");
if (mysql_num_rows($res) != 0) {
$array[]= array("starcount" => "already entered");
} else {
$sql = mysql_query("INSERT INTO Review(book_id,starcount,user_email) values('.$bookid.','.$starcount.','.$email')");
$nex = mysql_query("SELECT * FROM Review WHERE book_id='$bookid'");
while($row = mysql_fetch_array($nex)){
$star = $row["starcount"];
$array[] = array("starcount" => $star);
}
}
echo json_encode($array);
//echo "[{"name":"n1","city":"NY"},{"name":"n2","city":"Paris"}, ...]
?>

Answer 1

It seems to me "book_id" in "Review" table is primary key, as you have tried to add it multiple time, system shows the error duplicate key. Check the error & post it. Also check whether insert query is working or not.

Answer 2

you should not pass the primary key value manually
try this it will helps you
<?php
mysql_connect("localhost","root","");
mysql_select_db("FunReads");
$starcount="2";
$email = "vinodh#gmail.com";
$res=mysql_query("SELECT * FROM Review WHERE email ='$email'");
if(mysql_num_rows($res)!=0){
$array[]= array("starcount" => "already entered");
}else{
$sql=mysql_query("INSERT INTO Review (starcount,email) values('.$starcount.','.$email')");
$nex=mysql_query("SELECT * FROM Review WHERE email='$email'");
while($row=mysql_fetch_array($nex)){
$star = $row["starcount"];
$array[] = array("starcount" => $star);
}
}
echo json_encode($array);
?>

Answer 3

I just updated your code and it is working fine for me.
<?php
mysql_connect("localhost","user","");
mysql_select_db("xyz");
$email = "hari#gmail.com";
$starcount = "2";
$bookid = "5";
$sql = "SELECT * FROM review WHERE book_id='$bookid' AND user_email='$email'";
$res = mysql_query($sql);
if (mysql_num_rows($res) != 0) {
$array[]= array("starcount" => "already entered");
} else {
$sql = "INSERT INTO review(book_id,starcount,user_email) values('$bookid','$starcount','$email')";
$sql = mysql_query($sql);
$nex = mysql_query("SELECT * FROM review WHERE book_id='$bookid'");
while($row = mysql_fetch_array($nex)){
$star = $row["starcount"];
$array[] = array("starcount" => $star);
}
}
echo json_encode($array);
sample output :
[{"starcount":"2"},{"starcount":"3"},{"starcount":"1"},{"starcount":"2"},{"starcount":"1"}]
I updated the insert query, please try to update the same and test.