我有一个按钮，单击时会显示一个表单。它可以在我们的工作站上工作，也可以在云端工作，但不能在我的笔记本电脑上工作。其他项目在我的笔记本电脑上工作。
我收到这些错误：
警告：mysqli :: query（）：无法在第7行的G：\ Web Projects \ kmbgis \ assets \ caseForm.php中获取mysqli
致命错误：在第8行的G：\ Web Projects \ kmbgis \ assets \ caseForm.php中的非对象上调用成员函数fetch_assoc（）
这是 caseForm.php 中的PHP：
<?php
include('config.php');
//check any user action
$action = isset( $_POST['action'] ) ? $_POST['action'] : "";
$result = $conn->query("SHOW TABLE STATUS LIKE 'case_detail'");
$row = $result->fetch_assoc();
$nextCaseId = $row['Auto_increment'];
$result1 = $conn->query("SHOW TABLE STATUS LIKE 'suspect_data'");
$row1 = $result1->fetch_assoc();
$nextSuspectId = $row1['Auto_increment'];
$result2 = $conn->query("SHOW TABLE STATUS LIKE 'victim_data'");
$row2 = $result2->fetch_assoc();
$victimID = $row2['Auto_increment'];
//if the user hit the submit button on Case Detail
if($action == "Add"){
//Case Details Variables
$r_unit = $_POST['r_unit'];
$d_report = $_POST['d_report'];
$t_report = $_POST['t_report'];
$crimeCategory = $_POST['crime_category'];
$r_station = $_POST['r_station'];
if($crimeCategory == '101'){
$offenseType = $_POST['crime_type1'];
}else if($crimeCategory == '102'){
$offenseType = $_POST['crime_type2'];
}else if($crimeCategory == '103'){
$offenseType = $_POST['crime_type3'];
}else if($crimeCategory == '104'){
$offenseType = $_POST['crime_type4'];
}else if($crimeCategory == '105'){
$offenseType = $_POST['crime_type5'];
}else{
$offenseType = "NOT NULL";
}
//Insert Values
$sql = "INSERT INTO case_detail (ReportingUnit,DateOfReporting,TimeOfReporting,SuspectID,OffenseID,CrimeTypeID,VictimID,StationID)
VALUES('$r_unit','$d_report','$t_report','$nextSuspectId','$offenseType','$crimeCategory','$victimID','$r_station')";
if($conn->query($sql) === TRUE){
echo "<script>
window.location='https://localhost/kmbgis/assets/suspectArea.php?id=".$nextCaseId."';
</script>";
} else{
echo "Error: " . $sql . "<br>" . $conn->error;
}
}
$conn->close();
?>
单击按钮时，会在{em> casedetail.php 文件中插入<?php include'caseForm'; ?>。还有那个 casedetail.php 我有PHP代码打开连接，显示查询并关闭连接。
connect.php 包含：
<?php
$servername = "localhost";
$username = "wis1-29";
$password = "Password";
$dbname = "wis1-29";
//Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
//Check connection
if($conn->connect_errno){
echo "Failed to connect to MySQL: (" . $conn->connect_errno . ") " . $conn->connect_error;
}
?>
这是包含 caseForm.php 的 casedetails.php 。 出于某种原因，它也没有显示我的 footer.php 。
<!DOCTYPE html>
<html lang="en">
<head>
<meta name="viewport" content="width=device-width, initial-scale=1">
<title>Police Database System</title>
<link rel="stylesheet" href="css/bootstrap.min.css">
<link rel="stylesheet" type="text/css" href="css/main.css">
<script src="js/jquery-1.11.2.js"></script>
<script src="js/jquery-migrate.js"></script>
</head>
<body>
<?php include 'assets/header.php'; ?>
<section class="container-fluid">
<?php include 'assets/loginForm.php'; ?>
<div class="jumbotron">
<h3 class="text-center">Police Database System</h3>
<h4 class="text-center">Case Details</h4>
</div>
<div>
<?php
include('assets/config.php');
$nameSql = "SELECT
case_detail.Crime_Case_ID,
offense.OffenseDescription,
suspect_data.SuspectID,
suspect_data.LastName,
suspect_data.FirstName,
case_detail.ReportingUnit,
station.StationName,
station.StationID,
case_detail.DateOfReporting,
case_detail.TimeOfReporting,
crime.CrimeTypeDescription,
victim_data.vLastName,
victim_data.vFirstName,
victim_data.VictimID
FROM case_detail
INNER JOIN suspect_data
ON case_detail.SuspectID = suspect_data.SuspectID
INNER JOIN offense
ON case_detail.OffenseID = offense.OffenseID
INNER JOIN crime
ON case_detail.CrimeTypeID = crime.CrimeTypeID
INNER JOIN victim_data
ON case_detail.VictimID = victim_data.VictimID
INNER JOIN station
ON case_detail.StationID = station.StationID";
$nameResult = $conn->query($nameSql);
if ($nameResult->num_rows > 0) {
echo "<table class='table table-bordered'>
<thead>
<tr>
<th>Crime Case Number</th>
<th>Crime Commited</th>
<th>Suspect(s)</th>
<th>Reporting unit</th>
<th>Police Station</th>
<th>Date of reporting</th>
<th>Time of reporting</th>
<th>Crime Category</th>
<th>Victim(s)</th>
<th>Update Case</th>
<th>Delete Case</th>
</tr>
</thead>";
// output data of each row
while($row = $nameResult->fetch_assoc()) {
echo "<tbody>";
echo"<tr>
<td>".$row["Crime_Case_ID"]."</td>
<td>".$row["OffenseDescription"]."</td>
<td>".$row["LastName"].", ".$row["FirstName"]."</td>
<td>".$row["ReportingUnit"]."</td>
<td>".$row["StationName"]."</td>
<td>".$row["DateOfReporting"]."</td>
<td>".$row["TimeOfReporting"]."</td>
<td>".$row["CrimeTypeDescription"]."</td>
<td>".$row["vLastName"].", ".$row["vFirstName"]."</td>";
echo'<td><a href="assets/caseUpdate.php?id='.$row["Crime_Case_ID"].'&s_id='.$row["SuspectID"].'&v_id='.$row["VictimID"].'&station_id='.$row["StationID"].' " class="btn btn-default">Edit</a></td>';
echo'<td><a href="assets/caseDelete.php?id='.$row["Crime_Case_ID"].' " class="btn btn-danger">Delete</a></td>';
echo"</tr>
</tbody>";
}
echo "</table>";
} else {
echo "0 results";
}
$conn->close();
?>
<a class="addbtn btn btn-default" href="#" id="addCase">New Blotter Entry</a>
</div>
<div class="center formWrap" id="caseWrap">
<button type="button" class="btn btn-danger btn-sm addbtn">
<span class="glyphicon glyphicon-remove"></span> Close
</button>
<h2 class="text-center">Enter new case:</h2>
<?php include 'assets/caseForm.php'; ?>
</div>
</section>
<?php include 'assets/footer.php'; ?>
<script src="js/bootstrap.min.js"></script>
<script src="js/main.js"></script>
</body>
</html>
答案 0 :(得分：0)
如果没有看到你的连接设置很难说，假设在config.php中，但这个错误通常是因为连接。第二个非对象错误是因为没有获得导致第一个错误的连接。
你可以发布连接字符串吗？应该看起来像这样...
$conn = new mysqli("HOST","USER","PASSWORD","TABLE_NAME");
答案 1 :(得分：0)
我安装了wamp，一切正常！我在不同的启动上重新安装了相同的xampp版本，它没有工作，所以问题可能是mysql版本。我从5.5.9更新了PHP版本，但仍然没有工作，所以我尝试更新xampp中的mysql，但未能这样做。所以我决定下载更新版本的xampp。
没有工作的xampp版本是1.8.3，5.5.9 PHP和5.6.16 MYSQL，工作的Wamp有PHP 5.5.12和5.6.17 MYSQL