PHP / SQL查询抛出错误[关闭]

Question

if (!empty($_POST['like'])) {
    $sql = "UPDATE Post set lcount = lcount + '1'";
    $result=mysql_query($sql);
    if (!$result) {
    printf("Error: %s
", mysqli_error($con));
    exit();
}

echo "<form action='".$_SERVER['PHP_SELF']."' method='post'>";
echo "<input type = 'submit' value = 'like' name='like'/>";
echo "</form>"; 

Trying to add like button like FB but my query is throwing an error. Lost please help. It is not displaying any detail error. Just writes error on my page...But I think its the query... Is my query right?

Answer 1

Your "error" most likely lies in your query statement.

$sql = "UPDATE Post set lcount = lcount + '1'";

By wrapping it in single quotes('), you're casting your number 1 as a string. MySQL can't possibly know how to add a string to an integer.

Instead, pass the number itself and not the string representation:

$sql = "UPDATE Post set lcount = lcount + 1";