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Home/ Questions/Q 8453631
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Editorial Team
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Editorial Team
Asked: June 10, 20262026-06-10T11:48:56+00:00 2026-06-10T11:48:56+00:00

I have following code to use google images search API: google.load(‘search’, ‘1’); function searchComplete(searcher)

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I have following code to use google images search API:

google.load('search', '1');   
    function searchComplete(searcher) {
      // Check that we got results
      if (searcher.results && searcher.results.length > 0) {
        // Grab our content div, clear it.
        var contentDiv = document.getElementById('contentimg');
        contentDiv.innerHTML = '';

        // Loop through our results, printing them to the page.
        var results = searcher.results;
        for (var i = 1; i < results.length; i++) {
          // For each result write it's title and image to the screen
          var result = results[i];
          var imgContainer = document.createElement('div');



          var newImg = document.createElement('img');
          // There is also a result.url property which has the escaped version
          newImg.src = result.tbUrl;


          imgContainer.appendChild(newImg);

          // Put our title + image in the content
          contentDiv.appendChild(imgContainer);

The problem is, it gives me 3 image results. How to break a loop and show only the 1st one instead of 3 images?
if I change for (var i = 1; i < results.length; i++) to for (var i = 3; i < results.length; i++) it shows only one image, but image shown is the 3rd one and I need to show 1st one 🙂
Please advice

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-10T11:48:57+00:00Added an answer on June 10, 2026 at 11:48 am

    Don’t use a for loop at all. Just replace all instances of i with 0.

    google.load('search', '1');   
        function searchComplete(searcher) {
          // Check that we got results
          if (searcher.results && searcher.results.length > 0) {
            // Grab our content div, clear it.
            var contentDiv = document.getElementById('contentimg');
            contentDiv.innerHTML = '';
    
            var result = searcher.results[0];
    
            var imgContainer = document.createElement('div');
    
            var newImg = document.createElement('img');
            // There is also a result.url property which has the escaped version
            newImg.src = result.tbUrl;
    
            imgContainer.appendChild(newImg);
    
            // Put our title + image in the content
            contentDiv.appendChild(imgContainer);
    

    0 means the first item returned (almost all number sequences in programming start at 0!) so all other results will be ignored.

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