PHP MYSQL $row[$variable]
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PHP MYSQL $row[$variable]
提问by Chinmay
I am trying to work around with dynamic table creation and data fetching. I am trying to get the data using following code :
我正在尝试解决动态表创建和数据获取。我正在尝试使用以下代码获取数据：
$myQuery = "SELECT ".$col_name." FROM ".$tabname." WHERE sampleid='".$sid."'";
$result = mysql_query($myQuery);
$row = mysql_fetch_array($result);
echo "<br>".$row['$col_name'];
But, I am unable to get any data back. I checked printing the query and running it in php my admin and its working as I want. But I guess variable in array might not be working I guess. Please help me regarding the same. Thanks.
但是，我无法取回任何数据。我检查了打印查询并在 php my admin 中运行它，并且它按我的意愿工作。但我猜数组中的变量可能不起作用。请帮助我。谢谢。
The Whole loop looks something like this :
整个循环看起来像这样：
$myQuery = "SELECT * FROM information_schema.columns WHERE table_name = '$tabname'";
$re = mysql_query($myQuery);
while($row = mysql_fetch_array ($re)){
if(!empty ($row)){
$col_name = $row['COLUMN_NAME'];
$myQuery = "SELECT ".$col_name." FROM ".$tabname." WHERE sampleid='".$sid."'";
echo "<br>".$myQuery;
$reqq = mysql_query($myQuery);
$roww = mysql_fetch_array($reqq);
echo "<br>".$roww[$col_name];
}
}
回答by Fluffeh
You are fetching an array, not an assoc array. Use this:
您正在获取一个数组，而不是一个关联数组。用这个：
echo "<br>".$row[0];
Edit: Having looked a little more, this may not be correct. You can set fetch_array to return assoc arrays.
编辑：看了一点，这可能不正确。您可以设置 fetch_array 以返回关联数组。
You cannot parse variables through single quotes '
您不能通过单引号 ' 解析变量
echo $row['col_name']; // manually typed string.
or
或者
echo $row[$col_name];
or
或者
echo $row["$col_name"];
回答by Jonathan de M.
You tried that->
你试过了->
echo "<br>".$row[$col_name];
OR
或者
$myQuery = "SELECT ".$col_name." FROM ".$tabname." WHERE sampleid='".$sid."'";
$result = mysql_query($myQuery);
$row = mysql_fetch_assoc($result);
echo "<br>".$row[$col_name];
Cause like said @Fluffeh it's not a associative array
因为就像@Fluffeh 所说的，它不是关联数组
回答by Brayden
Have you tried looping through the results array as such:
您是否尝试过循环遍历结果数组：
while($row = mysql_fetch_array($result)) {
echo $row['col_name'];
echo "<br />";
}
回答by Conrad Lotz
I would suggest the following solution. Give the parameter $col_nameand AS Col_NameThen the $row['Col_Name'] can always be set to the parameter no matter what the value.
我会建议以下解决方案。设置参数$col_name和AS Col_Name随后$行[“COL_NAME”]总是可以被设置为参数无论什么价值。
$myQuery = "SELECT ".$col_name." AS Col_Name FROM ".$tabname." WHERE sampleid='".$sid."'";
$result = mysql_query($myQuery);
$row = mysql_fetch_array($result);
echo "<br>".$row['Col_Name'];
