此代码必须正常工作,但事实并非如此。当我在mysql workbench上使用echo $ sql查询时,它实际上将数据添加到数据库中。我认为有一些关于mysql_query方法的东西。已经有两天我正在努力解决这个问题,一点点的帮助会很棒。
这是代码
<?php
if ($_SERVER["REQUEST_METHOD"]=="POST") {
include('mysqli_connect.php');
$AracMarkasi = mysqli_real_escape_string($conn, $_POST['AracMarkasi']);
$AracPlakasi = mysqli_real_escape_string($conn, $_POST['AracPlakasi']);
$AracRengi = mysqli_real_escape_string($conn, $_POST['AracRengi']);
$AracModeli = mysqli_real_escape_string($conn, $_POST['AracModeli']);
$AracSahibiAdi = mysqli_real_escape_string($conn,
$_POST['AracSahibiAdi']);
$AracSahibiSoyadi = mysqli_real_escape_string($conn,
$_POST['AracSahibiSoyadi']);
$AracSahibiAdresi = mysqli_real_escape_string($conn,
$_POST['AracSahibiAdresi']);
$AracSahibiTC = mysqli_real_escape_string($conn,
$_POST['AracSahibiTC']);
$DosyaNO = mysqli_real_escape_string($conn, $_POST['DosyaNO']);
$IcraDairesiNO = mysqli_real_escape_string($conn,
$_POST['IcraDairesiNO']);
$Durum = mysqli_real_escape_string($conn, $_POST['Durum']);
$sql = "INSERT INTO deptcar.arac_tablosu (AracPlakası,AracMarkasi,AracModeli,AracRengi,AracSahibiAdi,AracSahibiSoyadi,AracSahibiTCKimlik,AracSahibiAdresi,AracDurum,DosyaNO,IcraDairesiNO) VALUES('$AracPlakasi','$AracMarkasi','$AracModeli','$AracRengi','$AracSahibiAdi','$AracSahibiSoyadi','$AracSahibiTC','$AracSahibiAdresi','$Durum','$DosyaNO','$IcraDairesiNO')";
echo "$sql";
$result = mysqli_query($conn,$sql);
echo "$result";
if(!$result)
{ die('Could not enter data: ' . mysql_error());}
else{echo "Entered data successfully\n";}
mysql_close($conn);
}
?>
这部分是输出
已连接
INSERT INTO deptcar.arac_tablosu (AracPlakası,AracMarkasi,AracModeli,AracRengi,AracSahibiAdi,AracSahibiSoyadi,AracSahibiTCKimlik,AracSahibiAdresi,AracDurum,DosyaNO,IcraDairesiNO) VALUES( '广告', '广告', 'YT', '', '大鼠高血压相关基因', '汞柱', 'DHG', 'GGD', 'SahaPersoneli', 'GFH', 'HDG')
无法输入数据: