为什么我得到一个表使用PHPmyadmin和mysql在PHP中不存在错误？

Question

Setup -> Using host gator , php, phpmyadmin, mysql:

PHP Code (in the file called ajaxTest.php):

<?php

$con=mysqli_connect("localhost","refinedc_dbadmin","password","refinedc_currency");

// Check connection
if (mysqli_connect_errno()) 
{
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
else
{



$result = mysqli_query($con,"SELECT * FROM ITEMS");

if (!$check1_res) {
    printf("Error: %s
", mysqli_error($con));
    exit();
}

echo '[';

while($row = mysqli_fetch_array($result)) 
{
  echo '{';
  echo '"ID":' . '"' . $row['ID'] . '",';
  echo '"YEAR":' . '"' . $row['YEAR'] . '",';
  echo '"QUANTITY":' . '"' . $row['QUANTITY'] . '",';
  echo '"DENOMINATION":' . '"' . $row['DENOMINATION'] . '",';
  echo '"TYPE":' . '"' . $row['TYPE'] . '",';
  echo '"COUNTRY":' . '"' . $row['COUNTRY'] . '",';
  echo '"COIN_NAME_OR_TITLE":' . '"' . $row['COIN_NAME_OR_TITLE'] . '",';
  echo '"COLLECTIBLE_METAL_ONE":' . '"' . $row['COLLECTIBLE_METAL_ONE'] . '",';
  echo '"COLLECTIBLE_METAL_TWO":' . '"' . $row['COLLECTIBLE_METAL_TWO'];
  echo '},';
}

echo ']';

}
mysqli_close($con);

?>

---

phpmyadmin table:

---

mysql access levels for user and database:

---

user is added to the database:

---

And when i hit the page this is the issue i get:

---

UPDATE - Changed the SQL to read SELECT * FROM items... Now get this when I try to hit the page (hits the error code and runs exit() but prints out no error:

---

Why am i receiving a table does not exist error? I have already shortened the name of the database and rebuilt a new one, but it still says it doesn't exist!

Answer 1

Here is the corrected code. I have added comment in the code itself.

<?php

$con=mysqli_connect("localhost","refinedc_dbadmin","password","refinedc_currency");

// Check connection
if (mysqli_connect_errno()) 
{
    echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
else
{

    $result = mysqli_query($con,"SELECT * FROM `items`");
    // Check if result is ok
    // if there is an error $result will be false and the error details is available in mysqli_error()
    if (!$result) {
        printf("Error: %s
", mysqli_error($con));
        exit();
    }

    // Use json_encode to convert data to json.
    // It is too tedious and error prone to convert data to json by hand.
    $data = array();
    while($row = mysqli_fetch_array($result)) 
    {
        //Store the result to array
        $data[]=array(
            "ID" => $row['ID'],
            "YEAR" => $row['YEAR'],
            "QUANTITY" => $row['QUANTITY'],
            "DENOMINATION" => $row['DENOMINATION'],
            "TYPE" => $row['TYPE'],
            "COUNTRY" => $row['COUNTRY'],
            "COIN_NAME_OR_TITLE" => $row['COIN_NAME_OR_TITLE'],
            "COLLECTIBLE_METAL_ONE" => $row['COLLECTIBLE_METAL_ONE'],
            "COLLECTIBLE_METAL_TWO" => $row['COLLECTIBLE_METAL_TWO']
        );
    }
    //echo the json
    echo json_encode($data);
}
mysqli_close($con);

BTW: don't use W3schools if you really want to learn and do it in the correct way. Lot of the things there are outdated and misleading.