I'm Unable to set value of textbox using php variable

Question

I'm trying to set value of html input type text textboxes to empty when user clicks Search button and empID is not matched, but its giving error:

mysqli_num_rows() expects parameter 1 to be mysqli_result

Here is code:

<html>
<body>
<form action="" method="post">
<h2>Employee Form</h2>

<input type="text" name="empID">
<input type="submit" name="searchRec" value="Search" />
<hr>

  Employee ID: <input type="text" name="empIDC" value="<?php echo htmlentities($employeeID); ?>">
  <br><br>

  Name: <input type="text" name="name" value="<?php echo htmlentities($Name); ?>">
  <br><br>

  Address: <input type="text" name="address" value="<?php echo htmlentities($Address); ?>">
  <br><br>
</form>
<?php
if( isset( $_REQUEST['searchRec'] ))
{
$employeeID = ($_POST["empID"]);

$servername = "localhost";
$username = "root";
$password = "";
$dbname = "bc140_DB";

// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
    die("Connection failed: " . mysqli_connect_error());
}
$sql = "SELECT empID, Name, Address, Dateofbirth, Salary, Timein from Employee where empID == $employeeID";
$result = mysqli_query($conn, $sql);

if(mysqli_num_rows($result > 0)){   while($row = mysqli_fetch_array($result, MYSQL_ASSOC)) {   $employeeID = $row['empID'];   $Name = $row['Name'];   $Address = $row['Address'];   $Dateofbirth = $row['Dateofbirth'];   $Salary = $row['Salary'];   $timestamp = $row['timeIn']; } }else{     $employeeID = "";   $Name = ""; $Address = "";   $Dateofbirth = "";   $Salary = "";   $timestamp = ""; }

}
?>
</body>
</html>

Suggestion: use `isset()` in your inputs values to check and make sure the value is set before declaring it be placed there in the input; `value=""` — dale landry, Jul 08 '17 at 10:40
Possible duplicate of [PHP: "Notice: Undefined variable", "Notice: Undefined index", and "Notice: Undefined offset"](https://stackoverflow.com/questions/4261133/php-notice-undefined-variable-notice-undefined-index-and-notice-undef) — user3942918, Jul 08 '17 at 10:48

JYoThI · Accepted Answer · 2017-07-08T11:11:14.017

2

1st : Change your code order otherwise you will get undefined error . your trying the embed the variable with html before creating the variable .

2nd : should be use single = not == empID = $employeeID

3rd : your mixing mysql with mysqli here mysql_fetch_array($result, MYSQL_ASSOC)

Change to

mysqli_fetch_array($result,MYSQLI_ASSOC);

4th: And also use isset() to confirm that variable exists or not if exists echo it otherwise echo the empty string .

5th: change your if like this if(mysqli_num_rows($result)>0){ }

file.php

<?php
if( isset( $_REQUEST['searchRec'] ))
{
   ......
  $employeeID = $row['empID'];
  $Name = $row['Name'];
  $Address = $row['Address'];
  $Dateofbirth = $row['Dateofbirth'];
  $Salary = $row['Salary'];
  $timestamp = $row['timeIn'];
   ......
}
?>

<html>
<body>
.....
  Employee ID: <input type="text" name="empIDC" value="<?php  if(isset($employeeID)){ echo htmlentities($employeeID); } else { echo ""; }  ?>">
.....

</body>
</html>

edited Jul 08 '17 at 11:11

answered Jul 08 '17 at 10:32

JYoThI

11,977
1
11
26

I changed the order by making php code before but errors are still there. – Jazib_Prince Jul 08 '17 at 10:36
I changed code order, used = to match empID, and used mysqli_fetch_array($result,MYSQLI_ASSOC); but errors are still there – Jazib_Prince Jul 08 '17 at 10:46
can you update your new code in post . @Jazib_Prince – JYoThI Jul 08 '17 at 10:49
Thanks JYoThl by follwoing all steps you mentioned+using isset solved the issue. Now I'm trying to do this: if(mysqli_num_rows($result > 0)){ while($row = mysqli_fetch_array($result, MYSQL_ASSOC)) { $employeeID = $row['empID']; $Name = $row['Name']; $Address = $row['Address']; $Dateofbirth = $row['Dateofbirth']; $Salary = $row['Salary']; $timestamp = $row['timeIn']; } }else{ $employeeID = ""; $Name = ""; $Address = ""; $Dateofbirth = ""; $Salary = ""; $timestamp = ""; }its not working error: mysqli_num_rows() expects parameter 1 to be mysqli_result – Jazib_Prince Jul 08 '17 at 10:58
Please dont' push the code in comment box . it's too hard to read . try to edit your post using edit button @Jazib_Prince – JYoThI Jul 08 '17 at 11:04
change your if like this if(mysqli_num_rows($result)>0){ } – JYoThI Jul 08 '17 at 11:10
@JYoThl I edited the post and if(mysqli_num_rows($result)>0) this didn't worked – Jazib_Prince Jul 08 '17 at 11:12
Let us [continue this discussion in chat](http://chat.stackoverflow.com/rooms/148678/discussion-between-jazib-prince-and-jyothi). – Jazib_Prince Jul 08 '17 at 11:12
still your code not updated fully . here empID == $employeeID . please update entire code . @Jazib_Prince – JYoThI Jul 08 '17 at 11:15

score 1 · Answer 2 · answered Jul 08 '17 at 10:31

1

you have forgotten ';' value="<?php echo htmlentities($employeeID); ?>"

answered Jul 08 '17 at 10:31

Leo

7,274
5
26
48

Thanks for pointing out mistake, but errors are still there after inserting ; at the end of all statements. – Jazib_Prince Jul 08 '17 at 10:33
1

@Jazib_Prince and change the order like JYoThI said. – Leo Jul 08 '17 at 10:34
@Jazib_Prince try change ur query SELECT * FROM Employee where empID == $employeeID" and try to echo the variables after foreach loop to see if u are getting them – Leo Jul 08 '17 at 10:40
var_dump() your query to see if it gets results. Check the query stmt in SQL and see if passes an SQL entry in phpMyAdmin if available. – dale landry Jul 08 '17 at 10:42

I'm Unable to set value of textbox using php variable

2 Answers2