PHP: MySQLi-Query doesn't return true or false?

问题

If I run this code, the if-clause won't return false even though the data record doesn't exist in the database myDB in table myTable. I don't know what's wrong..

// MySQLi-Connection...

$result = $mysqli->query("SELECT * FROM `myDB`.`myTable` WHERE `itemtype` = 'comment' AND `itemID` = 3");
if ($result) {
echo "record found!";
} else {
echo "record not found!";
}

The record with itemID = 3 doesn't exist, but my if-clause says that the $result returns true..

回答1:

Returns FALSE on failure. For successful SELECT, SHOW, DESCRIBE or EXPLAIN queries
mysqli_query() will return a mysqli_result object. For other successful queries 
mysqli_query() will return TRUE.

#Reference PHP Manual

You can use

$result = $mysqli->query("YOUR QUERY")); 
if($result->num_rows){
 // Records Found
}else{
 // Empty Result
}

回答2:

Its much logical to use mysqli_num_rows here:

$result = $mysqli->query("SELECT * FROM `myDB`.`myTable` WHERE `itemtype` = 'comment' AND `itemID` = 3");
if($result->num_rows > 0) {
    echo 'record found';
} else {
    echo 'query is ok but no results found';
}

You can't put a mysqli_result object in your if since it will still evaluate as TRUE even if it yields 0 rows.

回答3:

Your PHP if statement will evaluate to TRUE, as an empty mysql results set does not mean it is a failed query. The query ran ok, and mysql retrieved 0 rows, hence $result evaluates to true.

To check if a valid row has been retrieved, try:

if ( ! empty($result) ) { 
  echo "record found!";
} else {
  echo "record not found!";
}

来源：https://stackoverflow.com/questions/24572271/php-mysqli-query-doesnt-return-true-or-false

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php

sql

mysqli