Question

我正在尝试制作一个PHP代码，如果我在数据库中放入一个URL，代码将会在该URL上找到它并查看它获得的HTTP代码但目前所做的只是给我HTTP - 代码0而不是怀特。

$username = "root";
$password = "luca170385";
$hostname = "localhost";

$dbhandle = mysqli_connect($hostname, $username, $password)
or file_put_contents($filename, date("Y-m-d H:i:s")."Unable to connect to database", FILE_APPEND);
echo "Connect to MySQL<br>";

$selected = mysqli_select_db($dbhandle , "accounts")
or file_put_contents($filename, date("Y-m-d H:i:s")."Could not select database<br>", FILE_APPEND);
echo "Selected Database<br>";

$sql = "SELECT `address` FROM `url` WHERE `code` IS NULL";

if ($result = mysqli_query($dbhandle, $sql)){

    while ($row = mysqli_fetch_assoc( $result));{
    printf ($row['address']);
    $output = "{$row['address']}";
    echo "{$row['address']}";
    $handle = curl_init($sql);
    curl_setopt($handle,  CURLOPT_RETURNTRANSFER, TRUE);

    $response = curl_exec($handle);

    $httpCode = curl_getinfo($handle, CURLINFO_HTTP_CODE);
    $update = "UPDATE url SET `code`= $httpCode WHERE `address` = $sql";
    echo "$update.<br>";
    echo "$handle.<br>";
    echo "$httpCode.<br>";

    mysqli_query($dbhandle, $update);
    echo "{$row['address']}";

    curl_close($handle);
    }
}else {
    echo 'Not working';
}

Answer 1

;会终止，因此它是空的。

while ($row = mysqli_fetch_assoc( $result));

因此，您只需删除;。

Answer 2

检查评论： -

<?php
error_reporting(E_ALL);// check all type of error
ini_set('display_errors',1);// display those errors
$username = "root";
$password = "luca170385";
$hostname = "localhost";

$dbhandle = mysqli_connect($hostname, $username, $password, "accounts");
/*or file_put_contents($filename, date("Y-m-d H:i:s")."Unable to connect to database", FILE_APPEND);
echo "Connect to MySQL<br>";*/// from where this $filename come from?

/*$selected = mysqli_select_db($dbhandle ,)
or file_put_contents($filename, date("Y-m-d H:i:s")."Could not select database<br>", FILE_APPEND);
echo "Selected Database<br>";*/ //not needed

$sql = "SELECT `address` FROM `url` WHERE `code` IS NULL";

if ($result = mysqli_query($dbhandle, $sql)){

    while ($row = mysqli_fetch_assoc( $result)){ // remove ;
        printf ($row['address']);
        $output = $row['address']; // remove {} and ""
        echo $row['address']; // remove {} and ""
        $handle = curl_init($sql);
        curl_setopt($handle,  CURLOPT_RETURNTRANSFER, TRUE);

        $response = curl_exec($handle);

        $httpCode = curl_getinfo($handle, CURLINFO_HTTP_CODE);
        $update = "UPDATE url SET `code`= $httpCode WHERE `address` = $output"; // not $sql
        echo "$update.<br>";
        echo "$handle.<br>";
        echo "$httpCode.<br>";
        mysqli_query($dbhandle, $update);
        echo $row['address'];// remove {} and ""
        curl_close($handle);
    }
}else {
    echo 'Not working';
}

为什么我可以让我的mysqli_fetch_assoc工作？

2 个答案: