assign selected href to a php variable

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assign selected href to a php variable

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the next function what it does is search for a typed word and show me the possible matches (search engine). When I show the matches I am doing it through a href and I would like to store the href in a php variable, to keep the searches that are made. My idea is to use the options where the user clicks, but there may be several options, not just one.

<?php 
if(!isset($_POST['search'])) exit('No se recibió el valor a buscar');

require_once 'conexion.php';

function search()
{
  $mysqli = getConnexion();
  $search = $mysqli->real_escape_string($_POST['search']);
  $query = "SELECT sintomas_resp FROM sintomas WHERE sintomas_resp LIKE '%$search%' ";
  $res = $mysqli->query($query);
  while ($row = $res->fetch_array(MYSQLI_ASSOC)) {
    echo "<p><a href='#'>$row[sintomas_resp]</a></p>";
  }
}

search();

php

asked by juancamilovallejos0 25.05.2018 в 17:21

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1 answer

help with this error: SyntaxError: JSON.parse: unexpected end of data at line 1 column 1 of the JSON data Separate PHP Array for BD insertion

score 0 · Accepted Answer

You just have to declare the variable out of the while and print it at the end ...

<?php 
if(!isset($_POST['search'])) exit('No se recibió el valor a buscar');

require_once 'conexion.php';

function search()
{
  $mysqli = getConnexion();
  $search = $mysqli->real_escape_string($_POST['search']);
  $query = "SELECT sintomas_resp FROM sintomas WHERE sintomas_resp LIKE '%$search%' ";
  $res = $mysqli->query($query);
  $Listado = "";
  while ($row = $res->fetch_array(MYSQLI_ASSOC)) {
    $Listado = $Liestado . "<p><a href='#'>$row[sintomas_resp]</a></p>";
  }
  echo ($Listado); 
}

search();