[英]PHP mySQLi query returning data but not displaying it
美好的一天,我已經對這個問題進行了廣泛的研究,但是不幸的是,沒有任何相關的問題解決了我的問題。
在這里,我有一個非常基本的PHP mySQLi db連接。 連接成功,在表上運行的查詢也將成功。 問題是結果集將不會顯示。 我所有的引用都是正確的,當我檢查是否填充了結果集時,它就是正確的。 我相信問題出在我的while塊上,但是運行時不會返回任何錯誤。 感謝您的時間
<?php
$db = mysqli_connect('localhost','root','','securitour') //connection to the database
or die('Error connecting to MySQL server.');
?>
<html>
<head>
</head>
<body>
<?php
$query = "SELECT * FROM location"; //The SQL query
mysqli_query($db, $query) or die('Error querying database.');
$result = mysqli_query($db, $query); //query the table an store the result set in a variable
$row = mysqli_fetch_array($result); //create an array and store the records of the result set in it
if (mysqli_num_rows($result) != 0) //to check if the result set contains data
{
echo "results found"; //THIS is what is returned.
}
else
{
echo "results not found";
}
while ($row = $result->fetch_assoc()) //itterate through the array and display the name column of each record
{
echo $row['name'];
}
mysqli_close($db);
?>
</body>
</html>
您不需要運行mysqli_query()
兩次。 並且您需要將mysqli_fetch_assoc
用於關聯數組
<?php
$db = mysqli_connect('localhost','root','','securitour') or die('Error connecting to MySQL server.');
?>
<html>
<head>
</head>
<body>
<?php
$query = "SELECT * FROM location"; //The SQL query
$result = mysqli_query($db, $query) or die('Error querying database.'); //query the table an store the result set in a variable
$row = mysqli_fetch_assoc($result); //create an array and store the records of the result set in it
if (mysqli_num_rows($result) != 0) //to check if the result set contains data
{
echo "results found"; //THIS is what is returned.
} else {
echo "results not found";
}
foreach ( $row as $name=>$val) {
echo $name . ':' . $val . '<br>';
}
mysqli_close($db);
?>
</body>
</html>
很多事情不就在這里 。
您要處理mysqli_query()函數兩次-不需要。
您正在選擇SQL查詢(SELECT *)中的所有字段。 您應該按名稱選擇字段。
嘗試以下方法:
<?php
$db = mysqli_connect('localhost','root','','securitour') //connection to the database
or die('Error connecting to MySQL server.');
?>
<html>
<head>
</head>
<body>
<?php
$query = "SELECT name FROM location"; //The SQL query
$result = mysqli_query($db, $query) or die('Error querying database'); //query the table an store the result set in a variable
if(mysqli_num_rows($result) > 0){
echo "Results found!";
while($row = mysqli_fetch_array($result)){ //create an array and store the records of the result set in it
echo $row['name'];
}
} else {
echo "results not found";
}
mysqli_close($db);
?>
</body>
</html>
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