Please note that a very old bug (#76548) has been fixed in 7.2.8.
Previously, pg_fetch_result did not fetch the next row if $row was omitted.
It is now well the case, so bad use of the function can now cause some bugs in your codes.
(PHP 4 >= 4.2.0, PHP 5, PHP 7, PHP 8)
pg_fetch_result — Returns values from a result instance
pg_fetch_result() returns the value of a particular row and field (column) in an PgSql\Result instance.
Note:
This function used to be called pg_result().
resultAn PgSql\Result instance, returned by pg_query(), pg_query_params() or pg_execute()(among others).
rowRow number in result to fetch. Rows are numbered from 0 upwards. If omitted, next row is fetched.
fieldA string representing the name of the field (column) to fetch, otherwise an int representing the field number to fetch. Fields are numbered from 0 upwards.
Boolean is returned as "t" or "f". All
other types, including arrays are returned as strings formatted
in the same default PostgreSQL manner that you would see in the
psql program. Database NULL
values are returned as null.
false is returned if row exceeds the number
of rows in the set, or on any other error.
| Version | Description |
|---|---|
| 8.3.0 |
row is now nullable.
|
| 8.1.0 |
The result parameter expects an PgSql\Result
instance now; previously, a resource was expected.
|
Example #1 pg_fetch_result() example
<?php
$db = pg_connect("dbname=users user=me") || die();
$res = pg_query($db, "SELECT 1 UNION ALL SELECT 2");
$val = pg_fetch_result($res, 1, 0);
echo "First field in the second row is: ", $val, "\n";
?>The above example will output:
First field in the second row is: 2
Please note that a very old bug (#76548) has been fixed in 7.2.8.
Previously, pg_fetch_result did not fetch the next row if $row was omitted.
It is now well the case, so bad use of the function can now cause some bugs in your codes.Comment on boolean fields:
If you retrieve a boolean value from the PostgreSQL database, be aware that the value returned will be either the character 't' or the character 'f', not an integer. So, the statement
if (pg_fetch_result($rsRecords,0,'blnTrueFalseField')) {
echo "TRUE";
} else {
echo "FALSE";
}
will echo "TRUE" in either case (True or False stored in the field). In order to work as expected, do this instead:
if (pg_fetch_result($rsRecords,0,'blnTrueFalseField') == 't') {
echo "TRUE";
} else {
echo "FALSE";
}See bug #33809 http://bugs.php.net/bug.php?id=33809
Whether this really is a bug or a feature is not clear.
However, it is probably best to always put your column names in extra quotes.
$res = pg_query(...);
$colname = pg_field_name($res, $j);
pg_fetch_result($res, $i, "\"$colname\"");In order to use upper case in pg_fetch_result column names, it is apparently necessary to include explicit quotation marks.
Thus when I do this sort of thing:
$res = pg_query(...);
$ncols = pg_num_fields($res);
for ($j = 0; $j < $ncols; ++$j) {
$colname[$j] = pg_field_name($res, $j);
$name = htmlspecialchars($colname[$j]);
print("Column $j name = \"$name\"\n");
$value = htmlspecialchars(pg_fetch_result($res, 0, $colname[$j]));
print("Column \"{$colname[$j]}\" value = \"$value\"\n");
}
I get this sort of thing:
[....]
Warning: pg_fetch_result() [function.pg-fetch-result]: Bad column offset specified in /.../view.php on line 247
Column 8 name = "VEC index"
Column "VEC index" value = ""
But if I change the $value line to this:
$value = htmlspecialchars(pg_fetch_result($res, 0, "\"$colname[$j]\""));
I get this:
[...]
Column 8 name = "VEC index"
Column "VEC index" value[0] = "47"
In my opinion, pg_fetch_result(...) should use the quotes already. In other words, this may be a bug in the PHP postgres library. It does not seem to be a documented feature of pg_fetch_result() although the postgresql manual documents it under "SQL syntax", "Lexical structure".
PHP version 5.1.4.
psql version 8.1.4.Use can use pg_fetch_result when getting a value (like a smallint as in this example) returned by your stored procedure
<?php
$pgConnection = pg_connect("dbname=users user=me");
$userNameToCheckFor = "metal";
$result = pg_query($pgConnection, "SELECT howManyUsersHaveThisName('$userNameToCheckFor')");
$count = pg_fetch_result($result, 0, 'howManyUsersHaveThisName');
?>