i'am trying to insert a JSON file into my MYSQL Database. But only the second line is inserted every time i run this script. along whit a bunch of errors for "Invalid argument supplied for foreach()"
I got this script from another webpage but i can't get it work smooth! Someone how can help me out?
php script:
<?php
$con = mysql_connect("localhost","***","***");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("***", $con);
$f = file_get_contents('http://.../data.json');
$arr = explode('},',$f); // Prepare for json_decode BUT last } missing
$global_arr = array(); // Contains each decoded json (TABLE ROW)
$global_keys = array(); // Contains columns for SQL
if(!function_exists('json_decode')) die('Your host does not support json');
for($i=0; $i<count($arr); $i++)
{
$decoded = json_decode($arr[$i].'}',true); // Reappend last } or it will return NULL
$global_arr[] = $decoded;
foreach($decoded as $key=> $value)
{
$global_keys[$key] = '';
}
}
// iterate $global_arr
for($i=0; $i<count($global_arr); $i++) // this is faster than foreach
{
// NOW use what ardav suggested
foreach($global_arr[$i] as $key => $value){
$sql[] = (is_numeric($value)) ? "`$key` = $value" : "`$key` = '" . mysql_real_escape_string($value) . "'";
}
$sqlclause = implode(",",$sql);
$rs = mysql_query("INSERT INTO temp_table SET $sqlclause");
} // for i
//
?>
the errors:
Warning: Invalid argument supplied for foreach() in ***.php on line 21
Warning: Invalid argument supplied for foreach() in ***.php on line 21
Warning: Invalid argument supplied for foreach() in ***.php on line 31
Warning: implode() [function.implode]: Invalid arguments passed in ***.php on line 34
Warning: Invalid argument supplied for foreach() in ***.php on line 31
Json file:
{
"data": [
{
"name": "name freiend 1",
"id": "friend id 1"
},
{
"name": "name freiend 2",
"id": "friend id 2"
},
{
"name": "name freiend 3",
"id": "friend id 3"
}
],
"paging": {
"next": "https://graph.facebook.com/100002295143005/friends?access_token=***"
}
}