[英]How can I stop my PHP from spamming the error.log with PHP Notice?
I'm currently working on a project where I need to get some data from 3 different tables out of a database and echo them so I can work with the results in jQuery. 我目前正在一个项目中,需要从数据库的3个不同表中获取一些数据并回显它们,以便可以在jQuery中处理结果。 I'm sending a GET-request with 3 variables to my PHP. 我正在向我的PHP发送一个带有3个变量的GET请求。 The first one is used to determine which 'command' needs to be executed, the other 2 are used to determine which table and which row needs to be queried. 第一个用于确定需要执行哪个“命令”,其他两个用于确定需要查询哪个表和哪个行。
This is the code I have so far: 这是我到目前为止的代码:
} elseif($_GET['command']=='getpage') {
$mypid = $_GET['id'];
$mytable = $_GET['table'];
$link = mysqli_connect($dbserver,$userdb,$passdb,$db_typo) or die(mysqli_error($link));
if($mytable == 'tableName1'){
$query = 'SELECT * FROM table1 WHERE uid = "'.$mypid.'"'; //I need 6 elements from this table
} elseif($mytable == 'tableName2'){
$query = 'SELECT * FROM table2 WHERE uid = "'.$mypid.'"'; //I need 7 elements from this table
} elseif($mytable =='tableName3'){
$query = 'SELECT * FROM table3 WHERE uid = "'.$mypid.'"'; //I need 8 elements from this table
} else {
echo 'no such table supported for this command';
}
$result = mysqli_query($link, $query) or die(mysqli_error($link));
$pagecontent = array();
while($row = mysqli_fetch_assoc($result)){
$pagecontent[] = array(
'id' => utf8_encode($row['uid']),
'name' => utf8_encode($row['name']),
'text1' => utf8_encode($row['text1']), //not every table has this
'text2' => utf8_encode($row['text2']),
'img' => utf8_encode($row['image']),
'parent' => utf8_encode($row['parent']), //not every table has this
'sub_parent' => utf8_encode($row['sub_parent']), //not every table has this
'deleted' => utf8_encode($row['deleted'])
);
}
echo '{"content": '.json_encode($pagecontent).'}';
}
I have over 50 pages which I need to get from the database. 我有超过50页需要从数据库中获取。 So when I would let the jQuery function that sends the GET-request run trough I would end up spamming the error.log with 因此,当我让发送GET请求的jQuery函数运行通过时,我最终会向error.log发送垃圾邮件,
PHP Notice: Undefined index: text1 in /var/www/my.php on line 171 PHP注意:未定义的索引:第171行的/var/www/my.php中的text1
which I don't want. 我不要
Is there another way to fix this 'problem' than just putting the query and while-loop inside the if-statement? 除了将查询和while循环放入if语句之外,还有另一种解决此“问题”的方法吗?
添加检查是否存在数组键
'text1' => isset($row['text1']) ? utf8_encode($row['text1']) : '',
I don't know whats the reason of why don't you want an if to check for indices, but you could add another loop inside the while
我不知道什么的,为什么你不想要的,如果检查指标,但你可以添加里面另一个循环的原因while
and add those variables: 并添加这些变量:
while($row = mysqli_fetch_assoc($result)){
$temp = array();
foreach($row as $key => $value) {
$temp[$key] = utf8_encode($value);
}
$pagecontent[] = $temp;
// $pagecontent[] = array_map('utf8_encode', $row);
}
echo json_encode(array('content' => $pagecontent));
This just simply gets all the contents inside your $row
, encodes all values then push it inside your container. 这只是简单地获取$row
所有内容,对所有值进行编码,然后将其推入容器中。
Sidenote: Don't build the JSON string by hand, create the final array structure, then encode in the end. 旁注:请勿手动构建JSON字符串,创建最终的数组结构,然后最后进行编码。
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