关于MySQLI语法的PHP文件导入错误

Question

我正在尝试将文件导入到数据库中，无论我进行什么更改，我都会遇到相同的错误。错误是 -

您在SQL语法中有错误；检查与您的MariadB服务器版本相对应的手册，以获取正确的语法，以便在第1行中使用" 1"

似乎找不到解决方案。我究竟做错了什么？谢谢：(

<?php
$conn = mysqli_connect('localhost','root');
if (!$conn) {
    die(mysqli_error());
} 
$db = mysqli_query($conn,"CREATE DATABASE IF NOT EXISTS monthly");
if (mysqli_query($conn,$db)){
    echo "Database created";
} else {
    echo "Database not created: " . mysqli_error($conn);
}
mysqli_select_db($conn, "monthly");
$ct = mysqli_query($conn,"CREATE TABLE IF NOT EXISTS `month1`(
`week1` INT(4) NOT NULL,
`week2` INT(4) NOT NULL,
`week3` INT(4) NOT NULL,
`week4` INT(4) NOT NULL
)");
if (mysqli_query($conn,$ct)){
    echo "Table created";
} else {
    echo "table not created: " . mysqli_error($conn);
}
$open = fopen('/xampp/htdocs/month1.txt','r');
while (!feof($open)) 
{
    $getTextLine = fgets($open);
    $explodeLine = explode(',',$getTextLine, 4);
    if(count($explodeLine) !=4) {
        continue;
    }
    $week1 = $explodeLine[0];
    $week2 = $explodeLine[1];
    $week3 = $explodeLine[2];
    $week4 = $explodeLine[3];
    list($week1,$week2,$week3,$week4) = $explodeLine;

    $qry = "insert into 'month1' ('week1','week2','week3','week4') values('$week1','$week2','$week3','$week4')" or die(mysqli_error());
    mysqli_query($conn,$qry);
}
fclose($open);
mysqli_close($conn);
?>

Answer 1

请参阅添加的评论，以解释我做过的修正案

<?php
// this needs a password, I assume yours in blank
$conn = mysqli_connect('localhost','root', '');
if (!$conn) {
    die(mysqli_error());
} 
// this creates a database
$status = mysqli_query($conn,"CREATE DATABASE IF NOT EXISTS monthly");
if ($status){
    echo "Database created";
} else {
    echo "Database not created: " . mysqli_error($conn);
}
mysqli_select_db($conn, "monthly");
$ct = mysqli_query($conn,"CREATE TABLE IF NOT EXISTS `month1`(
                        `week1` INT(4) NOT NULL,
                        `week2` INT(4) NOT NULL,
                        `week3` INT(4) NOT NULL,
                        `week4` INT(4) NOT NULL
                        )");
if ($ct){
    echo "Table created";
} else {
    echo "table not created: " . mysqli_error($conn);
}
$open = fopen('/xampp/htdocs/month1.txt','r');
while (!feof($open)) 
{
    $getTextLine = fgets($open);
    $explodeLine = explode(',',$getTextLine, 4);
    if(count($explodeLine) !=4) {
        continue;
    }
    $week1 = $explodeLine[0];
    $week2 = $explodeLine[1];
    $week3 = $explodeLine[2];
    $week4 = $explodeLine[3];
    list($week1,$week2,$week3,$week4) = $explodeLine;
    // Column and databse names are wrapped in backticks
    // Text data is wrapped in single quotes
    // integer data CAN be wrapped in single quote, but does not have to be
    $qry = "insert into `month1` (`week1`,`week2`,`week3`,`week4`) 
                        values($week1,$week2,$week3,$week4)" 
    // This is placing a query in a string variable 
    // so this die() is nonsense and anyway it meeds a parameter in the mysqli_error()
    // like this mysqli_error($conn)
    //   or die(mysqli_error());
    // this execues the query above
    mysqli_query($conn,$qry);
    // if you made the above line into 
    // $res = mysqli_query($conn,$qry);
    // you could check if it actually worked like this
    /*
    if ( !$res ) {
        mysqli_error($conn);
    }
    */
}
fclose($open);
mysqli_close($conn);
?>

请参阅MySQL中何时使用单语引号，双引号和背景。

Answer 2

我更改代码上部的几行，请检查和比较，有一些错误的错误，例如

if (mysqli_query($conn,$db)) and if (mysqli_query($conn,$ct)){ 
 the above lines have no meaning.

请在该地点或您的位置添加以下代码。

<?php
$conn = mysqli_connect('localhost','root','');
if (!$conn) {
    die(mysqli_error());
} 
$db = mysqli_query($conn,"CREATE DATABASE IF NOT EXISTS monthly");
if ($db){
    echo "Database created";
} else {
    echo "Database not created: " . mysqli_error($conn);
}
mysqli_select_db($conn, "monthly");
$ct = mysqli_query($conn,"CREATE TABLE IF NOT EXISTS `month1`(
`week1` INT(4) NOT NULL,
`week2` INT(4) NOT NULL,
`week3` INT(4) NOT NULL,
`week4` INT(4) NOT NULL
)");
if ($ct){
    echo "Table created";
} else {
    echo "table not created: " . mysqli_error($conn);
}