php - Mysqli query doesn't work twice
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I cannot get my Mysqli queries to both work. If I comment out one function in my html, the other function is properly executed and vice versa.
function all_posts() {
require_once 'database.inc.php';
$mysqli = mysqli_connect($host, $username, $password, $database);
$query = mysqli_query($mysqli, "SELECT variable_name, post_name, post_date, post_display FROM blog_posts ORDER BY id DESC LIMIT 5");
if (!$query)
echo mysqli_error();
while ($results = mysqli_fetch_assoc($query)) {
$post_name = $results['post_name'];
$post_date = $results['post_date'];
$post_display = $results['post_display'];
$variable_name = $results['variable_name'];
echo "<a href='posts.php?post={$variable_name}'>";
echo "<div class='entry'>";
echo "<div class='entry_header'>";
echo "<h2>{$post_name}</h2>";
echo "<h3>{$post_date}</h3>";
echo "</div>";
echo "<p>{$post_display}</p>";
echo "</div>";
echo "</a>";
}
mysqli_free_result();
}
function all_sidebar_posts() {
require_once 'database.inc.php';
$mysqli = mysqli_connect($host, $username, $password, $database);
$query = mysqli_query($mysqli, "SELECT variable_name, post_name FROM blog_posts ORDER BY id DESC LIMIT 5");
while ($results = mysqli_fetch_assoc($query)) {
$post_name = $results['post_name'];
$variable_name = $results['variable_name'];
echo "<li><a href='posts.php?post=$variable_name'>$post_name</a></li>";
}
mysqli_free_result();
}
Here is the html that I am outputting to.
<ul>
<?php all_sidebar_posts(); ?>
</ul>
</div>
<div class="content_container">
<?php all_posts(); ?>
</div>
I have tried usingmysqli_data_seek(); but haven't had luck. Perhaps I am not using it right? I have browsed many questions and found similar ones but I have tried them all to no avail. I am new to programming so I may be overlooking something basic. Thank you all for the help!
Answer
Solution:
You are doing it wrong way.
Never mix your data manipulation code with presentation code.
First, get the posts into array:
require_once 'database.inc.php';
mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
$mysqli = mysqli_connect($host, $username, $password, $database);
$sql = "SELECT variable_name, post_name, post_date, post_display
FROM blog_posts ORDER BY id DESC LIMIT 5"
$result = mysqli_query($mysqli, $sql);
$data = array();
while ($row = mysqli_fetch_assoc($result)) {
$data[] = $row;
}
and then use this$data array to display posts any times you need, simply usingforeach()
Answer
Solution:
http://www.php.net/manual/en/mysqli-result.data-seek.php
Consult the manual for the usage of data_seek();
Take this example:
$Query = "SELECT * FROM Users WHERE ID='1'";
$TheQuery -> $MySQLi->query($Query);
$Results = $TheQuery->fetch_array(MYSQLI_ASSOC);
$TheQuery->data_seek(0); // Lets you re-use the query
$Count = $TheQuery->num_rows; // Gets the count
so in your case:
You should perform the procedure method:
$query = "SELECT Name, CountryCode FROM City ORDER BY Name";
if ($result = mysqli_query($link, $query)) {
/* fetch row */
$row = mysqli_fetch_row($result);
printf ("City: %s Countrycode: %s\n", $row[0], $row[1]);
mysqli_data_seek($result, 0);
$row_cnt = mysqli_num_rows($result);
/* free result set*/
mysqli_free_result($result);
}
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