Insert data if number of rows is greater than 0 does not work

问题

I have this PHP function which will insert urls into MySQL, but only those which didn't exist until moment of executing. I have a problem, my if condition is simply ignored, so everything goes to database, ignoring the condition. Code is here:

function storeLink($url,$gathered_from) {

    global $conn;   
    $querycheck = "SELECT COUNT(*) FROM test WHERE link = '$url'";
    $resultcheck = mysqli_query($conn, $querycheck);
    $row = mysqli_fetch_array($resultcheck, MYSQLI_ASSOC);

    if($row['COUNT(*)'] < 1);
    {
       echo "<font color='red'>".$row['COUNT(*)']."</font>";
       $url = mysqli_real_escape_string($conn,$url);
       $gathered_from = mysqli_real_escape_string($conn,$gathered_from);
       $query = "INSERT INTO test (link, source) VALUES ('$url', '$gathered_from')";
       mysqli_query($conn,$query) or die('Error, insert query failed');
    }
}

I tried with MySQL count and also with PHP mysqli_num_rows, but still the same. No matter what the condition in if statement is, it simply ignores it. Help please...

回答1:

function storeLink($url,$gathered_from) {

    global $conn;   
    $querycheck = "SELECT COUNT(*) as CNT FROM test WHERE link = '$url'";
    $resultcheck = mysqli_query($conn, $querycheck);
    $row = mysqli_fetch_array($resultcheck, MYSQLI_ASSOC);

    if($row['CNT'] < 1) {
       echo "<font color='red'>".$row['CNT']."</font>";
       $url = mysqli_real_escape_string($conn,$url);
       $gathered_from = mysqli_real_escape_string($conn,$gathered_from);
       $query = "INSERT INTO test (link, source) VALUES ('$url', '$gathered_from')";
       mysqli_query($conn,$query) or die('Error, insert query failed');
    }
}

function storeLink($url,$gathered_from) {

    global $conn;   
    $querycheck = "SELECT link FROM test WHERE link = '$url'";
    $resultcheck = mysqli_query($conn, $querycheck);
    $row = mysqli_fetch_array($resultcheck, MYSQLI_ASSOC);

    if(mysqli_num_rows($resultcheck)==0) {
       echo "<font color='red'>".mysqli_num_rows($resultcheck)."</font>";
       $url = mysqli_real_escape_string($conn,$url);
       $gathered_from = mysqli_real_escape_string($conn,$gathered_from);
       $query = "INSERT INTO test (link, source) VALUES ('$url', '$gathered_from')";
       mysqli_query($conn,$query) or die('Error, insert query failed');
    }
}

回答2:

try this if($row['COUNT(*)'] < 1){ instead of if($row['COUNT(*)'] < 1);{

    function storeLink($url,$gathered_from) {

        global $conn;   
        $querycheck = "SELECT COUNT(*) as TotalRow FROM test WHERE link = '$url'";
        $resultcheck = mysqli_query($conn, $querycheck);
        $row = mysqli_fetch_array($resultcheck, MYSQLI_ASSOC);

        if($row['TotalRow'] == 0)
        {
           echo "<font color='red'>".$row['TotalRow']."</font>";
           $url = mysqli_real_escape_string($conn,$url);
           $gathered_from = mysqli_real_escape_string($conn,$gathered_from);
           $query = "INSERT INTO test (link, source) VALUES ('$url', '$gathered_from')";
           mysqli_query($conn,$query) or die('Error, insert query failed');
        }
    }

回答3:

INSERT INTO test (link, source) VALUES ('$url', '$gathered_from')
WHERE NOT EXISTS (
   SELECT COUNT(*) as TotalRow FROM test WHERE link = '$url'
) LIMIT 1;

As you can see you can do whole work with just 1 query

来源：https://stackoverflow.com/questions/19976555/insert-data-if-number-of-rows-is-greater-than-0-does-not-work

标签

php

mysql

if-statement

count

rows