Question

我正在尝试根据输入的漏洞和威胁从漏洞表中获取行数。但是我总是得到一个值None来代替漏洞数。

$link = mysqli_connect("localhost", "root", "sharmi@08", "MySQL56");

// Check connection
if ($link === false) {
    die("ERROR: Could not connect. " . mysqli_connect_error());
}

// Escape user inputs for security
$Vulnerability = mysqli_real_escape_string($link, $_POST['Vulnerability']);
$Threat = mysqli_real_escape_string($link, $_POST['Threat']);
/*$P_Vulnerability = mysqli_real_escape_string($link, $_POST['P_Vulnerability']);*/
/*$P_Threat = mysqli_real_escape_string($link, $_POST['P_Threat']);*/

$Threat_count = mysqli_real_escape_string($link, $_POST['Threat_count']);
$table = mysqli_real_escape_string($link, $_POST['Vul']);  
// attempt insert query execution

$sql = "INSERT INTO vuln_threat(Vulnerability, Threat) VALUES ('$Vulnerability', '$Threat') ";
$Vulnerability_count = mysqli("Select count(*) from vuln_threat");

$sql = "UPDATE vuln_threat SET Vulnerability_Count='$Vulnerability_count',Threat_Count='1' WHERE Vulnerability='".$Vulnerability."' AND Threat='".$Threat."'";

//$select_result = mysqli_query($query);

if ( mysqli_query($link, $sql, $query)) {    
    echo "New Records added successfully." ;      

} else {
    echo "Please add records." ;
}
// close connection
mysqli_close($link);

Answer 1

当以过程方式使用时，mysqli_query仅接受预定义的常量作为第3个参数。

http://php.net/manual/en/mysqli.query.php

如何使用PHP计算表中的行数

1 个答案: