未定义索引:___mysqli_ston
Undefined index: ___mysqli_ston
我正在尝试将一些 json 数据从文件解析到 mysql 数据库中。我的原始代码如下:
<?php
$response = array();
$res=array();
$json = file_get_contents('C:\Users\Richard\Desktop\test.json');
if($json!=null){
$decoded=json_decode($json,true);
//$decode= var_dump($decoded);
//$ss=$decode["array"];
//echo $decoded['number'];
if(is_array($decoded["configurationItems"]))
{
foreach($decoded["configurationItems"] as $configurationItems)
//for($i=0;$i>sizeof($decoded["configurationItems"]);$i++)
{
$configurationItemVersion=$configurationItems["configurationItemVersion"];
echo "<br />","configurationItemVersion:",$configurationItemVersion,"<br />";
$configurationItemCaptureTime=$configurationItems["configurationItemCaptureTime"];
echo "configurationItemCaptureTime:",$configurationItemCaptureTime,"<br />";
$configurationStateId=$configurationItems["configurationStateId"];
echo "configurationStateId:",$configurationStateId,"<br />";
$result = mysql_query("INSERT INTO configuration_item(configurationItemVersion,configurationItemCaptureTime,configurationStateId)
VALUES('$configurationItemVersion','$configurationItemCaptureTime','$configurationStateId')")or die("Insert Failed ".mysql_error());;
}// check if row inserted or not
if ($result) {
// successfully inserted into database
$response["code"] = 1;
$response["message"] = "successfully stored";
// echoing JSON response
echo json_encode($response);
} else {
// failed to insert row
$response["code"] = 2;
$response["message"] = "Oops! An error occurred.";
// echoing JSON response
echo json_encode($response);
}
}
}
?>
由于被贬低而失败;所以我没有改变错误处理程序来忽略这个(非常糟糕的做法),而是选择使用此处提供的方便的转换工具转换为 mysqli :https://github.com/philip/MySQLConverterTool
我运行上述代码的转换器,它生成了以下内容:
<?php
$response = array();
$res=array();
$json = file_get_contents('C:\Users\Richard\Desktop\test.json');
if($json!=null){
$decoded=json_decode($json,true);
//$decode= var_dump($decoded);
//$ss=$decode["array"];
//echo $decoded['number'];
if(is_array($decoded["configurationItems"]))
{
foreach($decoded["configurationItems"] as $configurationItems)
//for($i=0;$i>sizeof($decoded["configurationItems"]);$i++)
{
$configurationItemVersion=$configurationItems["configurationItemVersion"];
echo "<br />","configurationItemVersion:",$configurationItemVersion,"<br />";
$configurationItemCaptureTime=$configurationItems["configurationItemCaptureTime"];
echo "configurationItemCaptureTime:",$configurationItemCaptureTime,"<br />";
$configurationStateId=$configurationItems["configurationStateId"];
echo "configurationStateId:",$configurationStateId,"<br />";
$result = mysqli_query($GLOBALS["___mysqli_ston"], "INSERT INTO configuration_item(configurationItemVersion,configurationItemCaptureTime,configurationStateId)
VALUES('$configurationItemVersion','$configurationItemCaptureTime','$configurationStateId')")or die("Insert Failed ".((is_object($GLOBALS["___mysqli_ston"])) ? mysqli_error(
"]) : (($___mysqli_res = mysqli_connect_error()) ? $___mysqli_res : false)));;
}// check if row inserted or not
if ($result) {
// successfully inserted into database
$response["code"] = 1;
$response["message"] = "successfully stored";
// echoing JSON response
echo json_encode($response);
} else {
// failed to insert row
$response["code"] = 2;
$response["message"] = "Oops! An error occurred.";
// echoing JSON response
echo json_encode($response);
}
}
}
?>
根据 运行 此代码,我在标题中收到错误消息(未定义的索引:___mysqli_ston)并且不知道如何修复它,任何帮助将不胜感激。
ps 我正在使用 laravel 框架,如果它有所作为或打开其他解决方案。
我现在知道该错误与我没有数据库连接字符串这一事实有关,即 $GLOBALS["___mysqli_ston 是由生成器生成的。
据我了解,laravel 负责在其 mvc 架构中定义数据库连接,因此不需要重新定义。考虑到这一点,我的代码会是什么样子?
你一定是用这样的方式连接到 mysql
$con=mysqli_connect("localhost","my_user","my_password","my_db");
替换你的这一行
$result = mysqli_query($GLOBALS["___mysqli_ston"], "INSERT INTO configuration_item(configurationItemVersion,configurationItemCaptureTime,configurationStateId)
有了这个
$result = mysqli_query($con, "INSERT INTO configuration_item(configurationItemVersion,configurationItemCaptureTime,configurationStateId)
我正在尝试将一些 json 数据从文件解析到 mysql 数据库中。我的原始代码如下:
<?php
$response = array();
$res=array();
$json = file_get_contents('C:\Users\Richard\Desktop\test.json');
if($json!=null){
$decoded=json_decode($json,true);
//$decode= var_dump($decoded);
//$ss=$decode["array"];
//echo $decoded['number'];
if(is_array($decoded["configurationItems"]))
{
foreach($decoded["configurationItems"] as $configurationItems)
//for($i=0;$i>sizeof($decoded["configurationItems"]);$i++)
{
$configurationItemVersion=$configurationItems["configurationItemVersion"];
echo "<br />","configurationItemVersion:",$configurationItemVersion,"<br />";
$configurationItemCaptureTime=$configurationItems["configurationItemCaptureTime"];
echo "configurationItemCaptureTime:",$configurationItemCaptureTime,"<br />";
$configurationStateId=$configurationItems["configurationStateId"];
echo "configurationStateId:",$configurationStateId,"<br />";
$result = mysql_query("INSERT INTO configuration_item(configurationItemVersion,configurationItemCaptureTime,configurationStateId)
VALUES('$configurationItemVersion','$configurationItemCaptureTime','$configurationStateId')")or die("Insert Failed ".mysql_error());;
}// check if row inserted or not
if ($result) {
// successfully inserted into database
$response["code"] = 1;
$response["message"] = "successfully stored";
// echoing JSON response
echo json_encode($response);
} else {
// failed to insert row
$response["code"] = 2;
$response["message"] = "Oops! An error occurred.";
// echoing JSON response
echo json_encode($response);
}
}
}
?>
由于被贬低而失败;所以我没有改变错误处理程序来忽略这个(非常糟糕的做法),而是选择使用此处提供的方便的转换工具转换为 mysqli :https://github.com/philip/MySQLConverterTool
我运行上述代码的转换器,它生成了以下内容:
<?php
$response = array();
$res=array();
$json = file_get_contents('C:\Users\Richard\Desktop\test.json');
if($json!=null){
$decoded=json_decode($json,true);
//$decode= var_dump($decoded);
//$ss=$decode["array"];
//echo $decoded['number'];
if(is_array($decoded["configurationItems"]))
{
foreach($decoded["configurationItems"] as $configurationItems)
//for($i=0;$i>sizeof($decoded["configurationItems"]);$i++)
{
$configurationItemVersion=$configurationItems["configurationItemVersion"];
echo "<br />","configurationItemVersion:",$configurationItemVersion,"<br />";
$configurationItemCaptureTime=$configurationItems["configurationItemCaptureTime"];
echo "configurationItemCaptureTime:",$configurationItemCaptureTime,"<br />";
$configurationStateId=$configurationItems["configurationStateId"];
echo "configurationStateId:",$configurationStateId,"<br />";
$result = mysqli_query($GLOBALS["___mysqli_ston"], "INSERT INTO configuration_item(configurationItemVersion,configurationItemCaptureTime,configurationStateId)
VALUES('$configurationItemVersion','$configurationItemCaptureTime','$configurationStateId')")or die("Insert Failed ".((is_object($GLOBALS["___mysqli_ston"])) ? mysqli_error(
"]) : (($___mysqli_res = mysqli_connect_error()) ? $___mysqli_res : false)));;
}// check if row inserted or not
if ($result) {
// successfully inserted into database
$response["code"] = 1;
$response["message"] = "successfully stored";
// echoing JSON response
echo json_encode($response);
} else {
// failed to insert row
$response["code"] = 2;
$response["message"] = "Oops! An error occurred.";
// echoing JSON response
echo json_encode($response);
}
}
}
?>
根据 运行 此代码,我在标题中收到错误消息(未定义的索引:___mysqli_ston)并且不知道如何修复它,任何帮助将不胜感激。
ps 我正在使用 laravel 框架,如果它有所作为或打开其他解决方案。
我现在知道该错误与我没有数据库连接字符串这一事实有关,即 $GLOBALS["___mysqli_ston 是由生成器生成的。
据我了解,laravel 负责在其 mvc 架构中定义数据库连接,因此不需要重新定义。考虑到这一点,我的代码会是什么样子?
你一定是用这样的方式连接到 mysql
$con=mysqli_connect("localhost","my_user","my_password","my_db");
替换你的这一行
$result = mysqli_query($GLOBALS["___mysqli_ston"], "INSERT INTO configuration_item(configurationItemVersion,configurationItemCaptureTime,configurationStateId)
有了这个
$result = mysqli_query($con, "INSERT INTO configuration_item(configurationItemVersion,configurationItemCaptureTime,configurationStateId)