Correction of the PHP search code
Code:if(isset($_REQUEST["term"])){
Prepare a select statement
$sql = "SELECT * FROM products WHERE name LIKE ?“;
if($stmt = mysqli_prepare($link, $sql)){
Bind variables to the prepared statement as parameters
mysqli_stmt_bind_param ($stmt, “s”, $param_term)
Set of Parameters
$param_term = '%'.$_REQUEST["term"] .and %;
Attempt to execute the prepared statement
if(mysqli_stmt_execute($stmt)){
$result = mysqli_stmt_get_result ($stmt)
Check number of rows in the result set
if(mysqli_num_rows($result) > 0){
FETCH result rows as an associative array
while($row = mysqli_fetch_array($result, MYSQLI_ASSOC)){
$name = $row [name]
$photo = $row [‘photo’]
$id = $row [”id”]
$ali_url = $row [”ali_url”]
$cat = $row [”cat”]
$etsy_add = $row [”etsy_add”]
if($etsy_add==0) { $etsy_add_label = 'The goods are not added to Etsy'; }
Echo’
“$name”‘.$etsy_add_label’“$cat”“$ali_url”The View
‘;
The
by else{
Echo “No matches found”
The
by else{
“Error: Could not be able to execute $sql.“ .MySQL_error ($link )
The
The
Close statement
mysqli_stmt_close ($stmt)
The
The close connection
MySQL_close ($link )
-
1 day11 USD
567 3 0 1 day11 USD
Write help.
The error occurs because mysqli_prepare cannot prepare a sql request and returns false, and the mysqli_stmt_close function accepts a "prepared sql request", i.e. Examples of the class mysqli_stmt
In general, you need to rewrite a little logic of your code and then everything will work.
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